Mole fraction of the solute in a 1.00 molal aqueous solution is

- 0.0177
- 0.0344
- 1.77
- 0.1770

The number of moles of oxygen in 1L of air containing 21% oxygen by volume, under standard conditions, is

- 0.0093 mole
- 2.10 mole
- 0.186 mole
- 0.21 mole

Percentage of Se in peroxidase anhydrase enzyme is 0.5% by weight (at. weight = 78.4), then minimum molecular weight of peroxidase anhydrase enzyme is

- 1.568 X 10
^{3} - 15.68
- 2.168 X 10
^{4} - 1.568 X 10
^{4}

The total number of valence electrons in 4.2g of N_{3}^{–} ion is (N_{A} is the Avogadro’s number)

- 2.1N
_{A} - 4.2N
_{A} - 1.6N
_{A} - 3.2N
_{A}

The hydrated salt Na_{2}SO_{4}.nH_{2}O, undergoes 55% loss in mass on heating and becomes anhydrous. The value of n will be:

- 5
- 3
- 7
- 10

A partially dried clay mineral contains 8% water. The original sample contained 12% water and 45% silica. The% of silica in the partially dried sample is nearly:

- 50%
- 49%
- 55%
- 47%

One mole of a mixture of CO and CO_{2} requires exactly 20g of NaOH in solution for complete conversion of all the CO_{2} into Na_{2}CO_{3}. How much NaOH would it require for conversion into Na_{2}CO_{3}, if the mixture (one mole) is completely oxidised to CO_{2}?

- 60g
- 80g
- 40g
- 20g

100 mL of PH_{3} when decomposed produces phosphorus and hydrogen. The change in volume is:

- 50mL increase
- 500mL decrease
- 900mL decrease
- none of these

2 gm Iron pyrite (FeS_{2}) is burnt with O_{2} to form Fe_{2}O_{3} and SO_{2}. The mass of SO_{2} produced is (Fe=56, S=32, O=16)

- 2gm
- 2.13gm
- 4gm
- 4.26gm

Which of the following pairs of gases contains the same number of molecules:-

- 16g of O
_{2}and 14g of N_{2} - 8g of O
_{2}and 22g of CO_{2} - 28g of N
_{2}and 22g of CO_{2} - 32g of O
_{2}and 32g of N_{2}

How many moles of KMnO_{4} are needed to oxidize a mixture of 1 mole of each FeSO_{4} & FeC_{2}O_{4} in acidic medium

- 4/5
- 5/4
- 3/4
- 5/3

Oxygen contains 90% O^{16} and 10% O^{18}. Its atomic mass is

- 17.4
- 16.2
- 16.5
- 17

At S.T.P. the density of CCl_{4} vapour in g/L will be nearest to

- 6.84
- 3.42
- 10.26
- 4.57

The number of gram molecules of oxygen in 6.02 × 10^{24} CO molecules is

- 10gm molecules
- 5gm molecules
- 1gm molecules
- 0.5gm molecules

The mass of carbon present in 0.5 mole of K_{4}[Fe(CN)_{6}] is

- 1.8g
- 18g
- 3.6g
- 36g

The oxide of metal contains 40% by mass of oxygen. The percentage of chlorine in the chloride of the metal is

- 84.7
- 74.7
- 64.7
- 44.7

The empirical formula of an organic compound containing carbon and hydrogen is CH_{2}. The mass of one litre of this organic gas is exactly equal to that of one litre of N_{2}. Therefore, the molecular formula of the organic gas is

- C
_{2}H_{4} - C
_{3}H_{6} - C
_{6}H_{12} - C
_{4}H_{8}

A sample of pure compound is found to have *Na* = 0.0887 mole, *O *= 0.132 mole, *C* = 2.65 × 10^{22} atoms. The empirical formula of the compound is

- Na
_{2}CO_{3} - Na
_{3}O_{2}C_{5} - NaCO
- Na
_{0.0887}O_{0.132}C_{2.65×1022}

Volume occupied by one molecule of water (density = 1g cm^{-3}) is

- 9.0 x 10
^{-23}cm^{3} - 6.023 x 10
^{-23}cm^{3} - 3.0 x10
^{-23}cm^{3} - 5.5 x10
^{-23}cm^{3}

A mixture of methane and ethene in the molar ratio of x : y has a mean molar mass of 20. What would be the mean molar mass. if the gases are mixed in the molar ratio of y : x?

- 22
- 24
- 20.8
- 19

A sample of phosphorus that weights 12.4 g exerts a pressure 8 atm in a 0.821 litre closed vessel at 527°C. The molecular formula of the phosphorus vapour is:

- P
_{2} - P
_{4} - P
_{6} - P
_{8}

What volume of HCl solution of density 1.2 g/cm^{3} and containing 36.5% by weight HCl, must be allowed to react with zinc(Zn) in order to liberate 4.0 g of hydrogen?

- 333.33 mL
- 500 mL
- 614.66 mL
- None of these

What is the molar mass of diacidic organic Lewis base (B), if 12 g of chloroplatinate salt (BH_{2}PtCl_{6}) on ignition produced 5 gm residue of Pt?

- 52
- 58
- 88
- None of these

Equivalent weight of FeS2FeS2 in the half reaction, FeS2 → Fe2O3+SO2FeS2 → Fe2O3+SO2 is:

- M/10
- M/11
- M/6
- M/1

The equivalent weight of HCl in the given reaction is: K_{2}Cr_{2}O_{7}+14HCl → 2KCl+2CrCl_{3}+3Cl_{2}+H_{2}O?

- 16.25
- 36.5
- 73
- 85.1

Equivalent weight of H_{3}PO_{2} when it disproportionate into PH_{3} and H_{3}PO_{3} is:

- M
- M/2
- M/4
- 3M/4

In preparation of iron from haematite (Fe_{2}O_{3}) by the reaction with carbon

Fe_{2}O_{3} + C → Fe + CO_{2}

How much 80% pure iron could be produced from 120 kg of 90% pure Fe_{2}O_{3} ?

- 94.5 Kg
- 60.48 Kg
- 116.66 Kg
- 120 Kg

One of the following combinations illustrate the law of reciprocal proportions:

- N
_{2}O_{3}, N_{2}O_{4}, N_{2}O_{5} - NaCl, NaBr, NaI
- CS
_{2}, CO_{2}, SO_{2} - PH
_{3}, P_{2}O_{3}, P_{2}O_{5}

The formula of an acid is HXO_{2}. The mass of 0.0242 moles of the acid 1.657 g. What is the atomic weight of X?

- 35.5
- 28.1
- 128
- 19.0

A 6.85 g sample of the hydrates Sr(OH)_{2}.xH_{2}O is dried in an oven to give 3.13 g of anhydrous Sr(OH)_{2}. What is the value of x? (Atomic weights : Sr=87.60, O=16.0, H=1.0)

- 8
- 12
- 10
- 6

A solution of Na_{2}S_{2}O_{3} is standardized iodometrically against 0.167 g of KBrO_{3}. This process requires 50 mL of the Na_{2}S_{2}O_{3} solution. What is the normality of the Na_{2}S_{2}O_{3}?

- 0.2N
- 0.12N
- 0.72N
- 0.02N

The NH_{3} evolved due to complete conversion of N from 1.12 g sample of protein was absorbed in 45 mL of 0.4N HNO_{3}. The excess acid required 20 mL of 0.1 N NaOH. The % N in the sample is:

- 8
- 16
- 20
- 25

Rearrange the following (I to IV) in the order of increasing masses:

(I) 0.5 mole of O_{3}

(II) 0.5 gm atom of oxygen

(III) 3.011××10^{23} molecules of O_{2}(IV) 5.6 litre of CO_{2} at STP

- II<IV<III<I
- II<I<IV<III
- IV<II<III<I
- I<II<III<IV

Density of dry air containing only N_{2} and O_{2} is 1.15 g/L at 740 mm and 300 K. What is % composition of N_{2} by weight in the air?

- 78%
- 75.5%
- 70.02%
- 72.75%

The vapour density of a mixture containing NO_{2} and N_{2}O_{4} is 27.6. The mole fraction of N_{2}O_{4} in the mixture is:

- 0.1
- 0.2
- 0.5
- 0.8

What volume of 75% alcohol by weight (d=0.80g/cm^{3}) must be used to prepare 150 cm^{3} of 30% alcohol by weight (d=0.90g/cm^{3})?

- 67.5 mL
- 56.25 mL
- 44.44 mL
- None of these

Average atomic mass of magnesium is 24.31 AMU. This magnesium is compound of 79 mole% of _{24}Mg and remaining 21 mole % of _{25}Mg and _{26}Mg. Calculate mole% of _{26}Mg.

- 10
- 11
- 15
- 16

Calculate the % of free SO_{3} in oleum (a solution of SO_{3} in H_{2}SO_{4}) that is labelled 109% H_{2}SO_{4}.

- 40
- 30
- 50
- None

Suppose two elements X and Y combine to form two compounds XY_{2} and X_{2}Y_{3} when 0.05 mole of XY_{2} weighs 5 g while 3.011××10^{23} molecules of X_{2}Y_{3} weighs 85 g. The atomic masses of x and y are respectively:

- 20, 30
- 30, 40
- 40, 30
- 80, 60

100 mL of 10% NaOH (w/V) is added to 100 mL of 10% HCl (w/V). The resultant solution becomes:

- Alkaline
- Strongly Alkaline
- Acidic
- Neutral

1.44 gram of titanium (At. wt. = 48) reacted with excess of O_{2} and produce x gram of non-stoichiometric compound Ti_{1.44}O . The value of x is:

- 2
- 1.77
- 1.44
- None of these

For the reaction;

2Fe(NO_{3})_{3} + 3Na_{2}CO_{3} → Fe_{2}(CO_{3})_{3}+6NaNO_{3}

Initially if 2.5 mole of Fe(NO_{3})_{3} and 3.6 mole of Na_{2}CO_{3} is taken. If 6.3 mole of NaNO_{3} is obtained then % yield of given reaction is:

- 50
- 84
- 87.5
- 100

The impure 6 g of NaCl is dissolved in water and then treated with excess of silver nitrate solution. The weight of precipitate of silver chloride is found to be 14 g. The % purity of NaClNaCl solution would be:

- 95%
- 85%
- 75%
- 65%

25.4 of I_{2} and 14.2 g of Cl_{2} are made to react completely to yield a mixture of ICI and ICI_{3}. Calculate mole of ICI and ICI_{3} formed.

- 0.5, 0.2
- 0.1, 0.1
- 0.1, 0.3
- 0.3, 0.4

The vapour density of a mixture containing NO_{2} and N_{2}O_{4} is 38.3. Calculate the mole of NO_{2} in 100 g mixture.

- 0.437
- 0.347
- 0.557
- 0.663

Average molar mass of mixture of CH_{4} and C_{2}H_{4} present in the mole ratio a : b is 20 g mol^{-1}. If the mass ratio is reversed, what will be molar mass of mixture?

- 24 gram
- 42 gram
- 20 gram
- 15 gram

The haemoglobin from the red blood corpuscles of most mammals contains approximately 0.33% of iron by mass. The molar mass of haemoglobin as 67,200. The number of iron atoms in each molecule of haemoglobin is (atomic mass of iron=56):

- 2
- 3
- 4
- 5

The vapour density of a volatile chloride of a metal is 95 and the specific heat of the metal is 0.13 cal/g. The equivalent mass of the metal will be:

- 6
- 12
- 18
- 24

10 mL of hydrogen combines with 5 mL of oxygen to yield water. When 200 mL of hydrogen at STP is passed over heated Cuo, the CuO loses 0.144 g of its weight. Do these results correspond to the law of constant composition?

**Solution**

Common salt obtained from Clifton beach contained 60.75% chlorine while 6.40 g of a sample of common salt from Khewra mine contained 3.888 g of chlorine. Show that these data are in accordance with the law of constant composition.

**Solution**

Common salt from Clifton beach contains = 67.75% Cl_{2}

100g of salt = 60.75g of Cl_{2}

1g of salt = 60.75/100 = 0.6075g of Cl_{2}

6.40g of NaCl from Khewra mine = 3.888g of Cl_{2}

1g of NaCl from Khewra mine = 3.888/6.40 = 0.6075g of Cl_{2}

Thus, the weight of Cl_{2} in 1g of salt in bith the cases is same. Hence, the law of constant composition is verified.

3.2 g sulphur combines with 3.2 g of oxygen, to form a compound in one set of conditions. In another set of

conditions 0.8 g of sulphur combines with 1.2 g of oxygen to form another compound. State the law illustrated by these chemical combinations.

**Solution**

3.2g of S combines with 3.2g of O_{2}

1g of S combines with = 1g of O_{2}

0.8g of S combines with 1.2g of O_{2}

1g of S combines with = 1.2/0.8 = 1.5g of O_{2}

Thus the ratio of O_{2} in both cases which combines with fixed mass (1g) of S = 1:1.5 or 2:3, which is a simple whole number ratio. Hence the law of multiple proportion is verified.

1 g of oxygen combines with 0.1260 g of hydrogen to form H_{2}O. 1 g of nitrogen combines with 0.2160 g of hydrogen to form NH_{3} . Predict the weight of oxygen required to combine with 1 g of nitrogen to form an oxide.

**Solution**

In H_{2}O

0.126g of H_{2} combines with 1g of oxygen.

1g of H_{2} combines with = 1/0.126 = 7.936g of O_{2}

In NH_{3}

0.216g of H_{2} combines with 1g of N_{2}

1g of H_{2} combines with = 1/0.216 = 4.629g of N_{2}

Ratio of O:N in H_{2}O and NH_{3} which combines with a fixed wight of H2 (1g) = 7.936 : 4.629 = 1.72 : 1

So, wight of oxygen which combines with 1g of N should be 1.71g or its simple multiple.

KCl contains 52% of potassium, Kl contains 23.6% of potassium, and ICI contains 77.8% of iodine. Show that the above data is in agreement with the law of reciprocal proportions.

**Solution**

KCl = 52%K, 48%Cl

KI = 23.6%K, 76.4%I

ICl = 77.8%I, 22.2%Cl

In KCl

48g of Cl reacts with 52g of K

1g of Cl reacts with = 52/48 = 1.08g of K

In ICl

22.2g of Cl reacts with 77.8g of I

Ratio of K and I in KCL and ICL which reacts with fixed mass of Cl (1g) = 1.08 : 3.5 = 1 : 3.2

Ration of K and I in KI = 23.6 : 76.4 = 1 : 3.2

These two ratios are = 1 : 1

This prove the law of reciprocal proportions

What weight of sodium chloride would be decomposed by 4.9 g of sulphuric acid, if 6 g of sodium bisulphate (NaHSO_{4}) and 1.825 g of hydrogen chloride were produced in the reaction and the law of conservation of mass is true?

**Solution**

NaCl + H2SO4 ——> NaHSO4 + HCl

According to law of conservation of mass, the weight of reactants must be equal to weight of products.

Suppose mass of NaCl in reaction is x

So

x + 4.9 = 6 + 1.825

x = 7.825 – 4.9 = 2.925 = 3g (Approximately)

If the law of constant compositionis true, what weights of calcium, carbon, and oxygen are present in 1.5 g of calcium carbonate, if a sample of calcium carbonate from another source contains the following percentage composition: Ca = 40.0%; C = 12.0%, and O = 48.0%?

**Solution**

According to law, the samples of calcium carbonate from two sources must be identical i.e. their percentage composition must be same.

That’s why, the weights of Ca, C and O in 1.5g of CaCO_{3} are =>

Calcium(Ca) = (40 X 1.5)/100 = 0.60g

Carbon(C) = (12 X 1.5)/100 = 0.18g

Oxygen(O) = (48X1.5)/100 = 0.72g

An element forms two oxides containing, respectively, 50% and 40% by weight of the element. Show that these oxides illustrate the law of multiple proportions.

**Solution**

50 g of element combines with 50 g of oxygen. Therefore, I g of element will combine with 1 g of oxygen.

In the other oxide: 40 g of element will combine with 60 g of oxygen. Therefore, 1g of element will combine with 64/40 or 1.5 g of oxygen. Thus, the weights of oxygen that combine with I g of the element in the two oxides are in the ratio of 1:1.5 or 2:3 which is a simple ratio. This illustrates the law of multiple proportions.

Elements A and B combines to form three different compounds :

0.3g of A + 0.4g of B → 0.7g of compound X

18.0g of A + 48.0g of B → 66.0g of compound Y

40.0g of A + 159.99g of B → 199.99g of compound Z

Show that the law of multiple proportions is illustrated by the data given above.

**Solution**

Weights of B that combine with 1.0g of A in the compounds X, Y and Z are respectively.

0.4/0.3 = 1.33

48.0/18.0 = 2.66

159.99/40.0 = 4.00

Ratio being 1.33 : 2.66 : 4.00 which comes out to be 1:2:3 simple ratio. This illustrates the law of multiple proportions.

An impure sample of sodium chloride that weighed 0.50g gave 0.90g of silver chloride as precipitate on treatment with excess of silver nitrate solution. Calculate the percentage purity of the sample.

**Solution**

NaCl + AgNO_{3} → AgCl + NaNO_{3}

Total weight of Reactants = NaCl + AgNO_{3} = 23 + 35.5 = 58.5g

Total weight of Products = AgCl + NaNO_{3} = 108 + 35.5 = 143.5g

143.5g of AgCl is obtained from 58.5g of pure NaCl.

0.9g of AgCl is obtained from = (58.5 X 0.9)/143.5 = 0.36g of pure NaCl

Percentage of purity = (0.36 X 100)/0.5 = 72%

How much magnesium sulphide can be obtained from 2.00g of Mg and 2.00g of S by the reaction.

Mg + S → MgS. Which is the limiting regeant? Calculate the amount of one of the reactants which remains unreacted?

**Solution**

Mg + S → MgS

32g of S is obtained from 24g of Mg

2g of S in obtained from = (24/32) X 2 = 1.5g of Mg

So, S is completely consumed and Mg is left

Hence, S is the limiting reagent

Therefore, the amount of product i.e MgS will be determined from S and not from Mg.

32g of S = 56g of MgS

2g of S = (56/32) X 2 = 3.5g of MgS

Amount of Mg unreacted = 2 – 1.5 = 0.5g

1.00 g of a hydrated salt contains 0.2014g of iron, 0.1153 g of sulphur, 0.2301g of oxygen and 0.4532g of water of crystallisation Find its empirical formula. (Fe = 56; S = 32; O = 16)

**Solution**

Element | Moles | Least Ratio |
---|---|---|

Fe | 0.2014/56 = 0.0036 | 0.0036/0.0036 = 1 |

S | 0.1153/32 = 0.0036 | 0.0036/0.0036 = 1 |

O | 0.2301/16 = 0.0143 | 0.0143/0.0036 = 3.97 = 4 |

H_{2}O | 0.4532/18 = 0.025 | 0.025/0.0036 = 6.94 = 7 |

An inorganic substance has the following compositions N=35%; H=59; O=60%

On being heated, it yielded a gaseous compound containing N = 63.63% and O = 36.37%. Suggest a formula for each substance and equation for the chemical change.

**Solution**

Element | Moles | Least Ratio | Whole no. Ratio |
---|---|---|---|

N | 15/14 = 2.5 | 2.5/2.5 = 1 | 2 |

H | 5/1 = 5 | 5/2.5 = 2 | 4 |

O | 60/16 = 3.75 | 3.75/2.5 = 1.5 | 3 |

Empirical formula will be N_{2}H_{4}O_{3} which can be written as NH_{4}NO_{3}

NH_{4}NO_{3} on heating will undergo through following reaction =>

NH_{4}NO_{3} → N_{2}O + 2H_{2}O

Element | Moles | Least Ratio |
---|---|---|

N | 63.63/14 = 4.545 | 4.545/2.273 = 1.99 = 2 |

O | 36.37/16 = 2.273 | 2.273/2.273 = 1 |

Empirical formula = N_{2}O

A compound of carbon, hydrogen, and nitrogen contains the three elements in the respective ratio of 9 : 1 : 3.5

Calculate the empirical formula. If the molecular weight of the compound is 108, what is its molecular formula?

**Solution**

C : H : N = 9 : 1 : 3.5

Moles Ration of C : H : N in compound = (9/12) : (1/1) : (3/5/14) = 0.75 : 1 : 10.25 = 3 : 4 : 1

Empirical formula = C_{3}H_{4}N

Empricial formula weight will be = C_{3}H_{4}N = 12 X 3 + 1 X 4 + 14 = 54

In order to molecular weight to be 108, emipirical formula need to be multiplied by 2. So molecular formula will be C_{6}H_{8}N_{2}.

045 g of an organic compound containing only C, H and N on combustion gave of 1.1g of CO_{2}, and 0.3 g of H_{2}O What is the percentage of C, H, and N in the organic compound.

**Solution**

% of C = (12/44) X ((Weight of CO2 X 100)/Weight of Compound)

% of C = (12 X 1.1 X 100)/(44 X 0.45) = 66.66%

% of H = (2/18) X ((Weight of H_{2}O X 100)/(Weight of Compound))

% of H = (2 X 0.3 X 100)/(18 X 0.45) = 7.4%

% of N = 100 – (66.66 + 7.4) = 25.93%

A pure sample of cobalt chloride weighing 130g was found to contain 0.59 g cobalt and 0.71 g chloride on quantitative analysis. What is the percentage composition of cobalt chloride.

**Solution**

1.39g of Compound contains = 0.59g of Co

100g of Compound contains = (0.59 X 100)/1.3 = 45.4%

1.30g of Compound = 0.71g of chloride

100g of Compound = (0.71 X 100)/1.3 = 54.6%

Glucose is a physiological sugar. What is the mass %C, mass %H and mass %O in glucose C_{6}H_{12}O_{6}.

**Solution**

Molecular weight of glucose C_{6}H_{12}O_{6} = 12 X 6 + 1 X 12 + 16 X 6 = 180g

180g of glucose = 72g of C

% of C in glucose = (72 X 100)/180 = 40% of C

% of H in glucose = (12 X 100)/180 = 6.67% of H

% of O in glucose = (96 X 100)/180 = 53.33% of O

So glucose have 40%, 6.67% and 53.33% of Carbon, Hydrogen and Oxygen respectively.

Find the weight of NaOH in its 50 milli equivalents.

**Solution**

Molecular weight of NaOH = 23 + 16 + 1 = 40g

1g equivalent of NaOH = 1000mEq = 40g

50mEq = (40 X 50)/1000 = 2g

Find the normality of H_{2}SO_{4} having 50 milli equivalents in 2 litres.

**Solution**

Normality = milli equivalent/volume of solution (in mL)

Normality = 50/2000 = 0.025N

Find the weight of H_{2}SO_{4}, in 1200 ml. of a solution of 0.2N strength.

**Solution**

Molecular weight of H_{2}SO_{4} = 2 X 1 + 32 + 16 X 4 = 98g

Equivalent weight of H_{2}SO_{4} = 98/2 = 49g

Normality = (Weight of H_{2}SO_{4} in Solution X 1000)/(Equivalent weight of H_{2}SO_{4}) X Volume of Solution (in mL)

0.2 = Weight of H_{2}SO_{4} in Solution X 1000 / 48 X 1200

Solving Above Equation we get **Weight of H _{2}SO_{4} in Solution** = 11.76g

What weight of Na_{2}CO_{3} of 95% purity would be required to neutralise 45.6 ml of 0.335N acid?

**Solution**

milli equivalents of Na_{2}CO_{3} neutralised = milli equivalents of acid used – **First Equation**

milli equivalents of Na_{2}CO_{3} = (weight of Na_{2}CO_{3} in solution ) X 1000/Equivalent weight of Na_{2}CO_{3}

milli equivalents of acid = Volume of acid used (in mL) X Normality of acid

milli equivalents of Na_{2}CO_{3} = weight of Na_{2}CO_{3} in solution X 1000/53

milli equivalents of acid = 45.6 X 0.235

Putting both of these together in **First Equation**

weight of Na_{2}CO_{3} X 1000/53 = 45.6 X 0.235

Solving above equation will give weight of Na_{2}CO_{3} in solution = 0.5679g**Finding Weight of Na _{2}CO_{3} in Solution if purity of solution is 95%**

Molecular weight of Na

_{2}CO

_{3}= 23 X 2 + 12 + 16 X 3 = 106g

Equivalent weight of Na

_{2}CO

_{3}= 106/2 = 53g

Weigth of Na

_{2}CO

_{3}in 95% pure solution = 0.5679 X 100/95 = 0.5978g

What is the strength in gram per litre of a solution of H_{2}SO_{4}, 12mL of which neutralised by 15mL of N/10 NaOH Solution?

**Solution**

milli equivalents of H_{2}SO_{4} in solution = milli equivalents of NaOH in solution

Normality of H_{2}SO_{4} Solution X Volume of H_{2}SO_{4} Solution = Normality of NaOH Solution X Volume of NaOH Solution

Normality of H_{2}SO_{4} Solution X 12 = 0.1 X 15

Solving above equation; **Normality of H _{2}SO_{4} Solution = 0.125**

Strenght of H

_{2}SO

_{4}in H

_{2}SO

_{4}Solution = Normality of H

_{2}SO

_{4}Solution X Equivalent Weight of H

_{2}SO

_{4}

Strenght of H

_{2}SO

_{4}in H

_{2}SO

_{4}Solution = 0.125 X 49 = 6.125g per litre

Two litres of NH_{3} at 30 ^{0}C and 0.20 atmosphere is neutralised by 134 mL of a solution of H_{2}SO_{4}, Calculate the normality of H_{2}SO_{4}.

**Solution**

Pressure X Volume = mole X R X Temperature

mole = PV/RT = 0.2 X 2 / 0.0821 X 303 = 0.01608 mol = 16.08 milli mol = 16.08 milli equivalents

Normality X Volume = milli equivalents

N X 134 = 16.08

N = 0.12

1g of calcium was burnt in excess of O_{2} and the oxide was dissolved in water to make up 1 L solution. Calculate the normality of alkaline solution.

**Solution**

Ca + 0.5O_{2} → CaO

40g of Ca = 56g of CaO

1g of Ca = 56/40 = 1.4g of CaO

Normality = (Weight of CaO X 1000)/ milli equivalents X volume (in mL)

Normality = 1.4 X 1000 / (56/2) X 1000 = 0.05N

What volume of a solution of hydrochloric acid containing 73g acid per litre would suffice for the exact neutralisation of sodium hydroxide obtained by allowing 0.46g of metallic sodium to act upon water.

**Solution**

milli equivalents of Na = milli equivalents of NaOH = milli equivalents of HCl

(0.46/23) X 1000 = (73/36.5) X V

Solving above equation **V = 10mL**

Find out the equivalent weight of H_{3}PO_{4} in the reaction:

Ca(OH)2 + H_{3}PO_{4} → CaHPO_{4} + 2H_{2}O

**Solution**

In the above equation, 2H atom has reacted so, H_{3}PO_{4} acts as a dibasic acid.

Equivalent weight of H_{3}PO_{4} = Molecular wight of H_{3}PO_{4} = 98/2 = 49g

(Molecular wight of H_{3}PO_{4} = 1 X 3 + 31 + 16 X 4 = 98)

What weight of AgCl will be precipitated when a solution containing 4.77g of NaCl is added to a solution of 5.77g of AgNO_{3}.

**Solution**

AgNO_{3} + NaCl → AgCl + NaNO_{3}

Reaction Stages | AgNO_{3} | NaCl | AgCl | NaNO_{3} |
---|---|---|---|---|

Initially | 5.77/170 = 0.033 | 4.77/58.5 = 0.08 | 0 | 0 |

After Reaction | 0 | 0.08 – 0.033 = 0.048 | 0.033 | 0.033 |

Weight to AgCl left after the reaction = 0.033 X 143.5 = 4.7355g

A sample of an alloy weighing 0.50g and containing 90% Ag was dissolved in concentrated HNO_{3}. Ag was analysed by Volhard method in which 25mL of KCNS was required for complete neutralisation. Determine the normality of KCNS.

**Solution**

Weight of Ag = (90 X 0.5)/100 = 0.45g

milli equivalents of Ag = milli equivalents of KCNS

(Weight of Ag) X 1000/(equivalents of Ag) = Volume in mL X Normality

0.45 X 1000/108 = 25 X N

Upon solving **N = 0.167**

Find the molality of H2SO4 solution whose specific gravity is 1.98g mL^{-1} and 95% by weight/volume H_{2}SO_{4}.

**Solution**

H_{2}SO_{4} is 95% by volume

Weight of H_{2}SO_{4} = 95g

Volume of solution = 100mL

Weight of solution = 100 X 1.98 = 198g

Weight of H_{2}O = 198 – 95 = 103g

molality of H_{2}SO_{4} = (Weight of H_{2}SO_{4}) X 1000 / (Weight of solution X Weight of H_{2}O )

molality of H_{2}SO_{4} = 95 X 1000 / 98 X 103 = 9.41

A 10 mL sample of human urine was found to have 5mg of urea on analysis. Calculate the molarity of the given sample.

**Solution**

Molarity of urine sample = (Weight of urea in sample ( in grams ) X 1000)/(Molecular wight of urea) X Volume of sample in mL

Weight of urea in sample (in grams) = 5 X 10^{-3}g

Molarity of urine sample = 5 X 10^{-3} X 1000 / 60 X 10 = 0.008

Molarity of urine sample = 0.008

Calculate the molarity and molality of 20% aqueous ethanol solution by volume given density of solution is 0.96g mL^{-1}

**Solution**

Solution is 20% by volume which means 20mL of C_{2}H_{5}OH is dissolved in 100mL of solution

Volume of H_{2}O = 100 – 20 = 80mL

Weight of H_{2}O = 80 X 1 = 80g (Density of water is 1)

Weight of solution = 100 X 0.96 = 96g

Weight of C_{2}H_{5}OH = 96 – 80 = 16g

Molarity = (Weight of C_{2}H_{5}OH X 1000)/(Molecular weight of C_{2}H_{5}OH) X 100

Molarity = 16 X 1000/46 X 100 = 3.48

Molality = (Weight of C_{2}H_{5}OH X 1000)/(Molecular weight of C_{2}H_{5}OH) X (Weight of H_{2}O)

Molality = 16 X 1000 / 46 X 80 = 4.35

If 4g NaOH is dissolved in 100mL of aqueous solution, what will be the difference in its normality and molarity?

**Solution**

Molarity = (Weight of NaOH X 1000)/(Molecular Weight of NaOH) X Volume of solution (in mL)

Molarity = 4 X 1000 / 40 X 100 = 1

Molarity = 1

As NaOH is a monobasic acid so its normality and molarity will be same. That’s why

Normality = 1

A solution contains 2.80 moles of acetone and 8.20 mole of CHCl_{3}. Calculate the mole fraction of acetone.

**Solution**

Mole fraction of acetone will be = 2.80 / (2.80 + 8.20) = 2.8 / 11 = 0.255

The percentage composition by weight of a solution is 45% X, 15% Y and 40% Z. Calculate the mole fraction of each component in the solution given molecular mass of X, Y, Z is 18, 60, 60 respectively.

**Solution**

Number of moles of X in compound = 45/18 = 2.5

Number of moles of Y in compound = 15/60 = 0.25

Number of moles of Z in compound = 40/60 = 0.666

Mole Fraction of X in compound = 2.5 / (2.5 + 0.25 + 0.666) = 0.73

Mole Fraction of Y in compound = 0.25 / (2.5 + 0.25 + 0.666) = 0.073

Mole Fraction of Z in compound = 0.666 / (2.5 + 0.25 + 0.666) = 0.195

What mass of (NH_{4})_{2}CO_{3} contains 0.4 mol NH_{4}^{–} ?

**Solution**

2 mol of NH_{4}^{+} = 1mol of (NH_{4})_{2}CO_{3} = 96g

0.4 mol NH_{4}^{+} = 96 X 0.4 / 2 = 19.2g

Mass of (NH4)_{2}CO_{3} which contains 6.03 X 10^{23} hydrogen atoms?

**Solution**

8 mol H atoms = 1 mol of (NH4)_{2}CO_{3} = 96g

1 mol of H atoms (6.03 X 10^{23}) = 96/8 = 12.0g

What volume of 98% H_{2}SO_{4} by mass, given density = 1.8g mL^{-1} , is required to prepare 3.0L of 3.0M H_{2}SO_{4}.

**Solution**

Molarity of solution = (Percentage of H_{2}SO_{4} in solution X 10 X d) / Molecular weight of H_{2}SO_{4}

Molarity of solution = 98 X 10 X 1.8 / 98 = 18M

(Before Dilution)Number of moles in 98% H_{2}SO_{4} = (After Dilution)Number of moles in 3.0L of 3.0M H_{2}SO_{4}*(Formula: number of moles = molarity X volume)*

18 X Volume of H_{2}SO_{4} needed = 3 X 3

Volume of H_{2}SO_{4} needed = 0.5 L = 500mL

How many mL of 3.0M HCl should be added to react completely with 21.0g of NaHCO_{3}.

HCl + NaHCO_{3} → NaCl + CO_{2} + H_{2}O

**Solution**

Moles of NaHCO_{3} = 21.0/84 = 0.25 mol

Moles of HCl = Moles of NaHCO_{3} *(Formuls => Number of moles = Molarity X Volume)*

3M X Volume of HCl = 0.25 mol

Volume of HCl = 83.3mL

What volume of 3.0M H_{2}SO_{4} is required to react with 6.54g of Zn.

Zn + H_{2}SO_{4} → ZnSO_{4} + H_{2}

**Solution**

Moles of Zn = 6.54/65.4 = 0.1 mol

Moles of H_{2}SO_{4} = moles of Zn*(moles = molarity X volume)*

3M X Volume of H_{2}SO_{4} = 0.1

Volume of H_{2}SO_{4} = 0.033L = 33.3 mL

Volume of H_{2}SO_{4} = 33.3 mL

How many mL of 0.5M KMnO_{4} solution will react completely with 92.0g of K_{2}C_{2}O_{4}.H_{2}O

16H^{–} + 2MnO_{4}^{+} + 5C_{2}O_{4}^{2-} → 10CO_{2} + 2Mn^{2+} + 8H_{2}O

**Solution**

Moles of K_{2}C_{2}O_{4}.H_{2}O = 92/184 = 0.5 mol

5 mol of C_{2}O_{4}^{2-} = 2 mol of MnO_{4}^{+}

0.5 mol of C_{2}O_{4}^{2-} = 0.2 mol of MnO_{4}^{+}

Molarity of KMnO_{4} solution **X** Volume of KMnO_{4} solution = 0.2 mol of MnO_{4}^{+}

0.5 X Volume of KMnO_{4} solution = 0.2

Volume of KMnO_{4} solution = 0.4L = 400mL

What volume of 8.0M H_{2}SO_{4} is needed to react completely with 108g of Al (Atomic Weight of Al = 27.0g)

2Al + 3H_{2}SO_{4} → Al_{2}(SO_{4})_{3} + H_{2}

**Solution**

Moles of Al = 108/27 = 4mol

Number of moles of H_{2}SO_{4} required for 4 mol Al = (3/2) X 4 = 6mol

Molarity X Volume of H_{2}SO_{4} = Number of moles of H_{2}SO_{4} required for 4 mol Al

8 X Volume of H_{2}SO_{4} = 6

Volume of H_{2}SO_{4} = 6/8 = 0.75L = 750mL

Molarity | Molarity depends on temperature, dilution and volume |

Molality | Molality depends only on dilution |

Mole Fraction | Mole fraction also depends only on dilution |

Normality | Normality like molarity depends on all |

How much Water will be produced if 1g of Hydrogen reacts with 1g of Oxygen react as per following reaction

2H_{2} + O_{2} → 2H_{2}O

**Solution**

1g of Hydrogen will be = 1/2 = 0.5 mol Hydrogen

1g of Oxygen will be = 1/32 = 0.03125 mol Oxygen

From reaction equation **2H _{2} + O_{2} → 2H_{2}O** its clear that Limiting Reagent would be

**Oxygen**. So amount of water being produced will depend upon on amount of oxygen being used.

As 1 mol of Oxygen produces 2 mol of Water (See Reaction Equation)

So => 0.03125 mol Oxygen will produce 0.03125 X 2 mol of Water = 0.0625 mol of Water.

Converting Water from Moles to gram => 0.0625 X 18 = 1.125gram of Water

**So 1g of Hydrogen and 1g of Oxygen upon reaction will produce 1.125g of Water**

How much Ammonia will be produced if 1g of Nitrogen and 1g of Hydrogen react as per following reaction

N_{2} + 3H_{2} → 2NH_{3}

**Solution**

1g of Nitrogen = 1/28 = 0.035 mol of Nitrogen

1g of Hydrogen = 1/2 = 0.5 mol of Hydrogen

From reaction equation **N _{2} + 3H_{2} → 2NH_{3}** its clear that limiting reagent would be

**Nitrogen**. So amount of Ammonia

being produced will only depends upon amount of Nitrogen used in the reaction.

As 1 mol of Nitrogen produces 2 mol of Ammonia (As per Reaction Equation)

Ammount of Ammonia produced from 0.035 mol of Nitrogen = 0.035 X 2 = 0.070 mol

Converting 0.070 mol of Ammonia to grams => 0.070 X 17 = 1.19gram of Ammonia

So amount of ammonia produced from 1g of Nitrogen and 1g of Hydrogen will be 1.19 gram only.

How much Calcium Oxide will be produced if 1g of Calcium Carbonate is burnt as per following reaction

CaCO_{3} → CaO + CO_{2}

**Solution**

As there is a single reactant only, so that should be **Limiting Reagent**.

As per equation 1 mol of Calcium Carbonate produces 1 mol of Calcium Oxide

1g of Calcium Carbonate = 1/100 = 0.01 mol

So 0.01 mol of Calcium Carbonate will produces 0.01 mol of Calcium Oxide

Converting 0.01 mol of Calcium Oxide to gram => 0.01 X 56 = 0.56 g

So 1g of Calcium Carbonate upon heating will produce 0.56g of Calcium Oxide.

How much methane will be produced if 1g of Carbon and 1g of Hydrogen reacts together as per following reaction

C + 2H_{2} → CH_{4}

**Solution**

As per the reaction equation, 1 mol of Carbon reacts with 2 mol of Hydrogen

So limiting reagent will be Carbon.

As per reaction equation, 1 mol of Carbon produces 1 mol of methane

So Ig of Carbon = 1/12 mol of Carbon will produce 1/12 mol of methane

Converting 1/12 mol of methane into grams => 1/12 X 16 = 16/12 = 1.33g of Methane

So, 1g of Carbon and 1g of Hydrogen produces 1.33g of Methane

Find Morality of 9.8% H_{2}SO_{4} by weight having density of 1.8g mL^{-1}

**Solution**

Molarity = ((Volume Percentage) X 10 X (Density of H_{2}SO_{4})) / Molecular weight of H_{2}SO_{4}

Molarity = 9.8 X 10 X 1.8 / 98 = 1.8M**Molarity = 1.8M**

Find number of molecules in 15.8g of KMnO_{4}

**Solution**

Number of moles in 15.8g of KMnO_{4} = 15.8 / 158 = 0.1 mol ( Molecular weight of KMnO_{4} = 158 )

Number of molecules = Number of moles X Avogadro Number

Number of molecules = 0.1 X 6.023 X 10^{23} = 6.023 X 10^{22}

So, Number of molecules in 15.8g of KMnO_{4} = 6.023 X 10^{22}

Calculate number of moles in 9.0g H_{2}C_{2}O_{4}

**Solution**

Number of moles = Given Mass / Molecular Mass of H_{2}C_{2}O_{4}

Molecular Mass of H_{2}C_{2}O_{4} = 1 X 2 + 12 X 2 + 16 X 4

Molecular Mass of H_{2}C_{2}O_{4} = 2 + 24 + 64 = 90**Molecular Mass of H _{2}C_{2}O_{4} = 90**

Number of moles = 9/90 = 0.1 mol of H

_{2}C

_{2}O

_{4}

Calculate number of moles and number of molecules in 8.8g of CO_{2}

**Solution**

Number of moles = Given Mass / Molecular Mass of CO_{2}

Molecular Mass of CO_{2} = 12 + 16 X 2 = 12 + 32 = 44

Number of moles = 8.8 / 44 = 0.2**Number of moles = 0.2**

Number of molecules = Number of moles X Avogadro Number

Number of molecules = 0.2 X 6.023 X 10^{23}**Number of molecules = 1.2046 X 10 ^{23}**

Which of the following pair of compounds illustrate the law of multiple proportions?

- SO
_{2}and SO_{3} - NO
_{2}and N_{2}O - MgO and Mg(OH)
_{2} - NO and N
_{2}O_{5}

**Solution**

SO_{2} and SO_{3} illustrates the law of multiple proportions here.

In a rocket motor fuelled with butane, 0.1 mol of butane requires what volume of O_{2} at STP for complete combustion.

**Solution**

Let’s see the reaction question of butane with oxygen

C_{4}H_{10} + 13/2 O_{2} → 4CO_{2} + 5H_{2}O

As per reaction equation 1 mol of C_{4}H_{10} reacts with 13/2 mol of oxygen

So 0.1 mol of C_{4}H_{10} reacts with 1.3/2 mol of oxygen

0.1 mol of C_{4}H_{10} reacts with 0.065 mol of oxygen

Converting moles of Oxygen to Litre => 0.065 X 22.4 = 1.456 Litre of Oxygen

(**Formula** => 1 mol of any gas always have volume of 22.4L)

A portable hydrogen generator utilises the following reaction

CaH_{2} + 2H_{2}O → Ca(OH)_{2} + 2H_{2}

2.1g of CaH2 would produce how much volume of H2 at STP.

**Solution**

As per Balanced Chemical Equation given above, 1 mol of CaH_{2} produces 2 mol of H_{2} meaning => 1 mol of CaH_{2} produces 2 X 22.4 litres of H_{2}**1 mol of CaH _{2} produces 2 X 22.4 litres of H_{2}**

Let’s convert 2.1g of CaH

_{2}to moles

Number of moles of CaH

_{2}= 2.1 / 22 = 0.095

0.095 mol of CaH

_{2}produces 0.095 X 2 X 22.4 = 4.256 litres of Hydrogen

Which of the following statements is/are wrong?

The following reactions occur:

P_{4} + 5O_{2} → P_{4}O_{10}

If 1.24g of P_{4} reacts with 8.0g of O_{2} then what amount of P_{4}O_{10} is produced.

**Solution**

As per reaction 1 mol P_{4} reacts with 5 mol O_{2} meaning => 124g of P_{4} reacts with 160g of O_{2}

So 1.24g of P_{4} reacts with 1.60g of O_{2} but 8.0g of O_{2} is given, that’s why for reaction Limiting Reagent would be P_{4}

1 mol of P_{4} produces 1 mol of P_{4}O_{10}

124g of P_{4} produces 444g of P_{4}O_{10}

=> 1.24g of P_{4} produces **4.44g of P _{4}O_{10}**

So amount of P

_{4}O

_{10}produced in the reaction is 4.44g

Which of the following statements is/are wrong?

The following reactions occur:

CS_{2} + 3Cl_{2 }→ CCl_{4} + S_{2}Cl_{2}

1.0g of CS_{2} and 2.0g of Cl_{2} reacts with each other, then what amount of CCl_{4} is formed?

**Solution**

As per reaction equation 1 mol of CS_{2} reacts with 3 mol of Cl_{2}

76g of CS_{2} reacts with 105g of Cl_{2}

But as per Question, only we have only 1.0g of CS_{2} and 2.0g of Cl_{2}

So 1.0g of CS_{2} reacts with 105/76 = 1.38g of Cl_{2} but we are given 2.0g of Cl_{2} . So for this reaction Limiting Reagent should be CS_{2}

1 mol of CS_{2} produces 1 mol of CCl_{4}

76g of CS_{2} produces 154g of CCl_{4}

1.0g of CS_{2} produces 154/76 = 2.02g of CCl_{4}

So amount of CCl_{4} produced in reaction is 2.02g

What amount of Al_{2}O_{3} is produced if 108g of Al and 213g of MnO is reacted as per following reaction

2Al + 3MnO → Al_{2}O_{3} + 3Mn

**Solution**

As per the reaction equation 2 mol of Al reacts with 3 mol of MnO

which means **54g of Al reacts with 261g of MnO**

According to reaction we have 108g of Al and 213g of MnO

So 108g of Al reacts with (261 X 108)54 = 522g of MnO

So as per balanced equation **108g of Al will require 522g of MnO** but we have only 213g of MnO, so **Al is a limiting reagent in the reaction**.

As per balanced reaction equation => 2 mol Al produces 1 mol of Al_{2}O_{3}

Which means => **54g of Al produces 102g of Al _{2}O_{3}**

108g of Al will produce (102 X 108)/54 = 204g of Al

_{2}O

_{3}

So in the reaction from 108g of Al and 213g of MnO only

**204g of Al**will be produced.

_{2}O_{3}Which reactant is the limiting reagent in following reaction, if 21.0g of Li reacts with 32.0g of O_{2}

4Li + O_{2} → 2Li_{2}O

**Solution**

As per balanced reaction equation, 4 mol of Li reacts with 1 mol of O_{2}**Meaning 28g of Li reacts with 32g of O _{2}**

Let’s see How much Oxygen is needed to react with 21.0g of Li

21.0g of Li reacts with (32 X 21)/28 = 24g of O

_{2}

So

**21.0g of Li actually required 24g of O**, but in the given reaction scenario 32.0g of O

_{2}to react with_{2}is given. That’s why for this reaction limiting reagent will be Lithium.

Which reactant is the limiting reagent in the following reaction, if 3.9g of K reacts with 4.26g of Cl_{2}

2K + Cl_{2} → 2KCl

**Solution**

As per balanced chemical equation 2 mol of K reacts with 1 mol of Cl_{2}

which means **39g of K reacts with 35g of Cl _{2}**

So, 3.9g of K will react with 3.5g of Cl

_{2}, but as per question there’s 4.26g of Cl

_{2}available. That’s why Limiting Reagent for the reaction will be K

Which reactant is the limiting reagent in the following reaction, if 106g of Na_{2}CO_{3} reacts with 109.5g of HCl

Na_{2}CO_{3} + 2HCl → 2NaCl + CO_{2} + H_{2}O

**Solution**

As per balanced chemical equation above => 1 mol of Na_{2}CO_{3} reacts with 2 mol of HCl

In other words, **108g of Na _{2}CO_{3} reacts with 37.5g of HCl**

So => 106g of Na

_{2}CO

_{3}will react with (37.5 X 106) / 108 = 36.80g of HCl

But as per question, there is 109.5g of HCl.

That’s why

**Na**.

_{2}CO_{3}would be Limiting Reagent27g of Al will react completely with ………………….g of O_{2} as per chemical equation 4Al + 3O_{2} → 2Al_{2}O_{3}

**Solution**

As per chemical equation => **4 mol of Al reacts with 3 mol of O _{2}**

108g of Al reacts with 96g of O

_{2}

So => 27g of Al will reacts with (96 X 27)/108 =

**24g of O**

_{2}Calculate equivalent weight of H_{3}PO_{4} in the following reaction

Ca(OH)_{2} + H_{3}PO_{4} → CaHPO_{4} + 2H_{2}O

**Solution**

As in the chemical reaction, only two of total three hydrogen atoms is replaced from H_{3}PO_{4} . So equivalent weight of H_{3}PO_{4} would be = Molecular Weight of H_{3}PO_{4} / 2 = 98/2 = 49g**Equivalent Weight of H _{3}PO_{4} in the reaction is 49g**.

When 10mL of ethyl alcohol having density = 0.7893g mL^{-1} is mixed with 20mL of water having density = 0.997g mL^{-1} at 25^{0}C, the final solution has density of 0.9571g mL^{-1}. Calculate the percentage change in total volume on mixing.

**Solution**

Total weight of alcohol and water = 10 X 0.7893 + 20 X 0.9971

Volume of mixture = (10 X 0.7893 + 20 X 0.9971) / 0.9571 = 29.08mL

Change in volume = (20 + 10) – 29.08 = 0.92mL

Percentage change in volume = (0.92 X 100) / 30 = 3.06%

The molality of 1L solution with x% H2SO4 is equal to 9. The weight of the solvent present in the solution is 910g. The value of x is =>

- 90
- 80.3
- 40.13
- 9

**Solution**

Molality = (1000 X Weight of Solute) / (Molecular Weight of H2SO4) X (Weight of Solvent)

9 = (1000 X Weight of Solute) / 98 X 910

Weight of Solute = (9 X 98 X 910) / 1000 = 802.62g L^{-1}

Density of 1M Solution of NaCl is 1.0585g mL^{-1}. Calculate molality of the solution.

**Solution**

Density of Solution = Molarity( (Molecular Weight of Solute/1000) + (1/molality of solution))

Putting values into the equation

1.0585 = 1 ((58.5/1000) + (1/molality of solution))

1.0585 = 0.0585 + 1/molality of solution

1/molality of solution = 1

So **Molality of Solution = 1**

If 0.5 mole of BaCl_{2} is mixed with 0.20 mol of Na_{3}PO_{4}, then find out maximum number of moles of Ba_{3}(PO_{4})_{2} which can be formed.

**Solution**

Balanced chemical equation of Reaction => 3BaCl_{2} + 2Na_{3}PO_{4} → 6NaCl + Ba_{3}(PO_{4})_{2}

So 3 mol of BaCl_{2} reacts with 2 mol of Na_{3}PO_{4}

Given that we have only 0.5 mol of BaCl_{2} So Na_{3}PO_{4} required to react with it will be (2/3) X 0.5 = 0.33 mol

So **As Per Balanced Chemical Equation**=> 0.5 mol of BaCl_{2} reacts with 0.33 mol of Na_{3}PO_{4}

But we are given only 0.20 mol of Na_{3}PO_{4}, that’s why in the reaction **Na _{3}PO_{4} will be Limiting Reagent**

That’s why amount of Ba

_{3}(PO

_{4})

_{2}formed from reaction will depends upon Na

_{3}PO

_{4}only.

As Per Balanced Chemical Equation =>

**2 mol of Na**

_{3}PO_{4}produces 1 mol of Ba_{3}(PO_{4})_{2}So 0.2 mol of Na

_{3}PO

_{4}will produce 0.2 mol of Ba

_{3}(PO

_{4})

_{2}

**So => Maximum number of moles of Ba**.

_{3}(PO_{4})_{2}which can be produced from chemical reaction between BaCl_{2}and Na_{3}PO_{4}is 0.2 molIf 0.5g of a mixture of two metals A and B displaces 560mL of H_{2} at STP from an acid. The Find out composition of the mixture. Given that Equivalent Weight of A, B = 12, 9 respectively.

**Solution**

1 mol of H_{2} = 22400mL = 2 Equivalents of H_{2}

1 Equivalent of H_{2} = 11200mL

Number of H_{2} Equivalents in 560mL of H_{2} = 560/11200 = 0.05**Number of H _{2} Equivalents in 560mL of H_{2} = 0.05**

As Per Rule of Number of Equivalents Being Same on Reactant Side and Products Side of a Chemical Equation.

**Equivalents of A + Equivalents of B = Equivalents of H**

_{2}Let x be Weight of A in mixture so weight of B in micture would be 0.5 – x

(Formula => Equivalents = Weight/Equivalent Weight)

Equivalents of A = x/12

Equivalents of B = (0.5 – x)/9

Putting all of these values in formula =>

**Equivalents of A + Equivalents of B = Equivalents of H**

_{2}x/12 + (0.5 – x)/9 = 0.05

Solving Above Equation we get

**x = 0.2**

So Weight of A in mixture = 0.2g and Weight of B in mixture = 0.5 – 0.2 = 0.3g

Percentage of A in mixture = (0.2/0.5) X 100 = 40%

Percentage of B in mixture = (0.3/0.5) X 100 = 60%

13.4g of a Sample of unstable hydrated salt Na_{2}SO_{4}.xH_{2}O was found to contain 6.3g of H_{2}O. The number of molecules of water of crystallisation is =>

- 5
- 7
- 2
- 10

**Solution**

Weight of Salt = 13.4g

Weight of H_{2}O = 6.3g

Weight of anhydrous salt = 13.4 – 6.3 = 7.1g

Moles of anhydrous salt = 7.1/842 = 0.05

Moles of H_{2}O = 6.3/18 = 0.35mol

0.05mol of anhydrous salt reacts with 0.35mol of H_{2}O

So, 1 mol of anhydrous salt reacts => 0.35/0.05 = 7 mol

5.6g of a metal forms 12.7g of metal chloride. Hence equivalent weight of the metal will be?

**Solution**

Weight of Cl reacted = 12.7 – 5.6 = 7.1g

7.1g of Cl reacts with 5.6g of metal

35.5g of Cl reacts with (5.6/7.1) X 35.5 = 28g of metal

So **Equivalent Weight of metal = 28g**

If molarity of H_{2}SO_{4} is 18M and its density is 1.8g mL^{-1} then what will be molality of H_{2}SO_{4}?

**Solution**

density = (molarity of H_{2}SO_{4} solution) X ((Molecular Weight of H_{2}SO_{4} / 1000) + (1/molality of H_{2}SO_{4}))

1.8 = 18(98/1000 + 1/molality of H_{2}SO_{4})

1.8 = 18(0.098 + 1/molality of H_{2}SO_{4})

0.1 = 0.098 + 1/molality of H_{2}SO_{4}

0.002 = 1/molality of H_{2}SO_{4}**molality of H _{2}SO_{4} = 500**

How many grams of phosphoric acid would be needed to neutralise 100g of magnesium hydroxide?

**Solution**

Molecular weight of Magnesium Hydroxide Mg(OH)_{2} = 58.3

Molecular weight of Phosphoric Acid H_{3}PO_{4} = 98

Chemical Equation of Reaction between Magnesium Hydroxide and Phosphoric Acid

3Mg(OH)_{2} + 2H_{3}PO_{4} → Mg_{3}(PO_{4})_{2} + 3H_{2}O

As Per Chemical Equation 3mol of Mg(OH)_{2} required for neutralising 2mol of H_{3}PO_{4} .

So technically 1mol of Mg(OH)_{2} can be neutralised with 2/3 mol of H_{3}PO_{4}.

So 1mol = 58.3g of Mg(OH)_{2} can be neutralised by 2/3 mol = (2/3) X 98 = 65.33g of H_{3}PO_{4}**=> 58.3g of Mg(OH) _{2} can be neutralised by 65.33g of H_{3}PO_{4}**

100g of Mg(OH)

_{2}can be neutralised by (65.33/58.3) X 100 = 112.05g of H

_{3}PO

_{4}

So,

**for neutralising 100g of Magnesium Hydroxide 112.05g of Phosphoric acid is needed**.