Basic Concepts of Chemistry Questions

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 103
  • 15.68
  • 2.168 X 104
  • 1.568 X 104

The total number of valence electrons in 4.2g of N3 ion is (NA is the Avogadro’s number)

  • 2.1NA
  • 4.2NA
  • 1.6NA
  • 3.2NA

The hydrated salt Na2SO4.nH2O, 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 CO2 requires exactly 20g of NaOH in solution for complete conversion of all the CO2 into Na2CO3. How much NaOH would it require for conversion into Na2CO3, if the mixture (one mole) is completely oxidised to CO2?

  • 60g
  • 80g
  • 40g
  • 20g

100 mL of PH3 when decomposed produces phosphorus and hydrogen. The change in volume is:

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

2 gm Iron pyrite (FeS2) is burnt with O2 to form Fe2O3 and SO2. The mass of SO2 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 O2 and 14g of N2 
  • 8g of O2 and 22g of CO2
  • 28g of N2 and 22g of CO2 
  • 32g of O2 and 32g of N2 

How many moles of KMnO4 are needed to oxidize a mixture of 1 mole of each FeSO4 & FeC2O4 in acidic medium

  • 4/5
  • 5/4
  • 3/4
  • 5/3

Oxygen contains 90% O16 and 10% O18. Its atomic mass is 

  • 17.4
  • 16.2
  • 16.5
  • 17

At S.T.P. the density of CCl4 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 × 1024 CO molecules is 

  • 10gm molecules
  • 5gm molecules
  • 1gm molecules
  • 0.5gm molecules

The mass of carbon present in 0.5 mole of K4[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 CH2. The mass of one litre of this organic gas is exactly equal to that of one litre of N2. Therefore, the molecular formula of the organic gas is

  • C2H4
  • C3H6
  • C6H12
  • C4H8

A sample of pure compound is found to have Na = 0.0887 mole, = 0.132 mole, C = 2.65 × 1022 atoms. The empirical formula of the compound is

  • Na2CO3
  • Na3O2C5
  • NaCO
  • Na0.0887O0.132C2.65×1022

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

  • 9.0 x 10-23 cm3
  • 6.023 x 10-23 cm3
  • 3.0 x10-23 cm3
  • 5.5 x10-23 cm3

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:

  • P2
  • P4
  • P6
  • P8

What volume of HCl solution of density 1.2 g/cm3 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 (BH2PtCl6) 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: K2Cr2O7+14HCl → 2KCl+2CrCl3+3Cl2+H2O?

  • 16.25
  • 36.5
  • 73
  • 85.1

Equivalent weight of H3PO2 when it disproportionate into PH3 and H3PO3 is:

  • M
  • M/2
  • M/4
  • 3M/4

In preparation of iron from haematite (Fe2O3) by the reaction with carbon
                          Fe2O3 + C → Fe + CO2
How much 80% pure iron could be produced from 120 kg of 90% pure Fe2O3 ?

  • 94.5 Kg
  • 60.48 Kg
  • 116.66 Kg
  • 120 Kg

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

  • N2O3, N2O4, N2O5
  • NaCl, NaBr, NaI
  • CS2, CO2, SO2
  • PH3, P2O3, P2O5

The formula of an acid is HXO2. 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.xH2O 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 Na2S2O3 is standardized iodometrically against 0.167 g of KBrO3. This process requires 50 mL of the Na2S2O3 solution. What is the normality of the Na2S2O3?

  • 0.2N
  • 0.12N
  • 0.72N
  • 0.02N

The NH3 evolved due to complete conversion of N from 1.12 g sample of protein was absorbed in 45 mL of 0.4N HNO3. 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 O3
(II)  0.5 gm atom of oxygen
(III) 3.011××1023 molecules of O2
(IV)  5.6 litre of CO2 at STP

  • II<IV<III<I
  • II<I<IV<III
  • IV<II<III<I
  • I<II<III<IV

Density of dry air containing only N2 and O2 is 1.15 g/L at 740 mm and 300 K. What is % composition of N2 by weight in the air?

  • 78%
  • 75.5%
  • 70.02%
  • 72.75%

The vapour density of a mixture containing NO2 and N2O4 is 27.6. The mole fraction of N2O4 in the mixture is:

  • 0.1
  • 0.2
  • 0.5
  • 0.8

What volume of 75% alcohol by weight (d=0.80g/cm3) must be used to prepare 150 cm3 of 30% alcohol by weight (d=0.90g/cm3)?

  • 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 24Mg and remaining 21 mole % of 25Mg and 26Mg. Calculate mole% of 26Mg.

  • 10
  • 11
  • 15
  • 16

Calculate the % of free SO3 in oleum (a solution of SO3 in H2SO4) that is labelled 109% H2SO4.

  • 40
  • 30
  • 50
  • None

Suppose two elements X and Y combine to form two compounds XY2 and X2Y3 when 0.05 mole of XY2 weighs 5 g while 3.011××1023 molecules of X2Y3 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 O2 and produce x gram of non-stoichiometric compound Ti1.44O . The value of x is:

  • 2
  • 1.77
  • 1.44
  • None of these

For the reaction; 
2Fe(NO3)3 + 3Na2CO3 → Fe2(CO3)3+6NaNO3
Initially if 2.5 mole of Fe(NO3)3 and 3.6 mole of Na2CO3 is taken. If 6.3 mole of NaNO3 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 I2 and 14.2 g  of Cl2 are made to react completely to yield a mixture of ICI and ICI3. Calculate mole of ICI and ICI3 formed.

  • 0.5, 0.2
  • 0.1, 0.1
  • 0.1, 0.3
  • 0.3, 0.4

The vapour density of a mixture containing NO2 and N2O4  is 38.3. Calculate the mole of NO2 in 100 g mixture.

  • 0.437
  • 0.347
  • 0.557
  • 0.663

Average molar mass of mixture of CH4 and C2H4 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% Cl2
100g of salt = 60.75g of Cl2
1g of salt = 60.75/100 = 0.6075g of Cl2
6.40g of NaCl from Khewra mine = 3.888g of Cl2
1g of NaCl from Khewra mine = 3.888/6.40 = 0.6075g of Cl2
Thus, the weight of Cl2 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 O2
1g of S combines with = 1g of O2
0.8g of S combines with 1.2g of O2
1g of S combines with = 1.2/0.8 = 1.5g of O2
Thus the ratio of O2 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 H2O. 1 g of nitrogen combines with 0.2160 g of hydrogen to form NH3 . Predict the weight of oxygen required to combine with 1 g of nitrogen to form an oxide.

Solution
In H2O
0.126g of H2 combines with 1g of oxygen.
1g of H2 combines with = 1/0.126 = 7.936g of O2

In NH3
0.216g of H2 combines with 1g of N2
1g of H2 combines with = 1/0.216 = 4.629g of N2

Ratio of O:N in H2O and NH3 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 (NaHSO4) 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 CaCO3 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 + AgNO3 → AgCl + NaNO3
Total weight of Reactants = NaCl + AgNO3 = 23 + 35.5 = 58.5g
Total weight of Products = AgCl + NaNO3 = 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

ElementMolesLeast Ratio
Fe0.2014/56 = 0.00360.0036/0.0036 = 1
S0.1153/32 = 0.00360.0036/0.0036 = 1
O0.2301/16 = 0.01430.0143/0.0036 = 3.97 = 4
H2O0.4532/18 = 0.0250.025/0.0036 = 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

ElementMolesLeast RatioWhole no. Ratio
N15/14 = 2.52.5/2.5 = 12
H5/1 = 55/2.5 = 24
O60/16 = 3.753.75/2.5 = 1.53

Empirical formula will be N2H4O3 which can be written as NH4NO3
NH4NO3 on heating will undergo through following reaction =>
NH4NO3 → N2O + 2H2O

ElementMolesLeast Ratio
N63.63/14 = 4.5454.545/2.273 = 1.99 = 2
O36.37/16 = 2.2732.273/2.273 = 1

Empirical formula = N2O


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 = C3H4N
Empricial formula weight will be = C3H4N = 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 C6H8N2.


045 g of an organic compound containing only C, H and N on combustion gave of 1.1g of CO2, and 0.3 g of H2O 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 H2O 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 C6H12O6.

Solution
Molecular weight of glucose C6H12O6 = 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 H2SO4 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 H2SO4, in 1200 ml. of a solution of 0.2N strength.

Solution
Molecular weight of H2SO4 = 2 X 1 + 32 + 16 X 4 = 98g
Equivalent weight of H2SO4 = 98/2 = 49g
Normality = (Weight of H2SO4 in Solution X 1000)/(Equivalent weight of H2SO4) X Volume of Solution (in mL)
0.2 = Weight of H2SO4 in Solution X 1000 / 48 X 1200
Solving Above Equation we get Weight of H2SO4 in Solution = 11.76g


What weight of Na2CO3 of 95% purity would be required to neutralise 45.6 ml of 0.335N acid?

Solution
milli equivalents of Na2CO3 neutralised = milli equivalents of acid used – First Equation
milli equivalents of Na2CO3 = (weight of Na2CO3 in solution ) X 1000/Equivalent weight of Na2CO3
milli equivalents of acid = Volume of acid used (in mL) X Normality of acid

milli equivalents of Na2CO3 = weight of Na2CO3 in solution X 1000/53
milli equivalents of acid = 45.6 X 0.235

Putting both of these together in First Equation
weight of Na2CO3 X 1000/53 = 45.6 X 0.235
Solving above equation will give weight of Na2CO3 in solution = 0.5679g

Finding Weight of Na2CO3 in Solution if purity of solution is 95%
Molecular weight of Na2CO3 = 23 X 2 + 12 + 16 X 3 = 106g
Equivalent weight of Na2CO3 = 106/2 = 53g
Weigth of Na2CO3 in 95% pure solution = 0.5679 X 100/95 = 0.5978g


What is the strength in gram per litre of a solution of H2SO4, 12mL of which neutralised by 15mL of N/10 NaOH Solution?

Solution
milli equivalents of H2SO4 in solution = milli equivalents of NaOH in solution
Normality of H2SO4 Solution X Volume of H2SO4 Solution = Normality of NaOH Solution X Volume of NaOH Solution
Normality of H2SO4 Solution X 12 = 0.1 X 15
Solving above equation; Normality of H2SO4 Solution = 0.125
Strenght of H2SO4 in H2SO4 Solution = Normality of H2SO4 Solution X Equivalent Weight of H2SO4
Strenght of H2SO4 in H2SO4 Solution = 0.125 X 49 = 6.125g per litre


Two litres of NH3 at 30 0C and 0.20 atmosphere is neutralised by 134 mL of a solution of H2SO4, Calculate the normality of H2SO4.

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 O2 and the oxide was dissolved in water to make up 1 L solution. Calculate the normality of alkaline solution.

Solution
Ca + 0.5O2 → 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 H3PO4 in the reaction:
Ca(OH)2 + H3PO4 → CaHPO4 + 2H2O

Solution
In the above equation, 2H atom has reacted so, H3PO4 acts as a dibasic acid.
Equivalent weight of H3PO4 = Molecular wight of H3PO4 = 98/2 = 49g
(Molecular wight of H3PO4 = 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 AgNO3.

Solution
AgNO3 + NaCl → AgCl + NaNO3

Reaction StagesAgNO3NaClAgClNaNO3
Initially 5.77/170 = 0.0334.77/58.5 = 0.0800
After Reaction00.08 – 0.033
= 0.048
0.0330.033

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 HNO3. 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 H2SO4.

Solution
H2SO4 is 95% by volume
Weight of H2SO4 = 95g
Volume of solution = 100mL
Weight of solution = 100 X 1.98 = 198g
Weight of H2O = 198 – 95 = 103g
molality of H2SO4 = (Weight of H2SO4) X 1000 / (Weight of solution X Weight of H2O )
molality of H2SO4 = 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 10-3g
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 C2H5OH is dissolved in 100mL of solution
Volume of H2O = 100 – 20 = 80mL
Weight of H2O = 80 X 1 = 80g (Density of water is 1)
Weight of solution = 100 X 0.96 = 96g
Weight of C2H5OH = 96 – 80 = 16g

Molarity = (Weight of C2H5OH X 1000)/(Molecular weight of C2H5OH) X 100
Molarity = 16 X 1000/46 X 100 = 3.48

Molality = (Weight of C2H5OH X 1000)/(Molecular weight of C2H5OH) X (Weight of H2O)
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 CHCl3. 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 (NH4)2CO3 contains 0.4 mol NH4 ?

Solution
2 mol of NH4+ = 1mol of (NH4)2CO3 = 96g
0.4 mol NH4+ = 96 X 0.4 / 2 = 19.2g


Mass of (NH4)2CO3 which contains 6.03 X 1023 hydrogen atoms?

Solution
8 mol H atoms = 1 mol of (NH4)2CO3 = 96g
1 mol of H atoms (6.03 X 1023) = 96/8 = 12.0g


What volume of 98% H2SO4 by mass, given density = 1.8g mL-1 , is required to prepare 3.0L of 3.0M H2SO4.

Solution
Molarity of solution = (Percentage of H2SO4 in solution X 10 X d) / Molecular weight of H2SO4
Molarity of solution = 98 X 10 X 1.8 / 98 = 18M

(Before Dilution)Number of moles in 98% H2SO4 = (After Dilution)Number of moles in 3.0L of 3.0M H2SO4
(Formula: number of moles = molarity X volume)
18 X Volume of H2SO4 needed = 3 X 3
Volume of H2SO4 needed = 0.5 L = 500mL


How many mL of 3.0M HCl should be added to react completely with 21.0g of NaHCO3.
HCl + NaHCO3 → NaCl + CO2 + H2O

Solution
Moles of NaHCO3 = 21.0/84 = 0.25 mol
Moles of HCl = Moles of NaHCO3 (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 H2SO4 is required to react with 6.54g of Zn.
Zn + H2SO4 → ZnSO4 + H2

Solution
Moles of Zn = 6.54/65.4 = 0.1 mol
Moles of H2SO4 = moles of Zn
(moles = molarity X volume)
3M X Volume of H2SO4 = 0.1
Volume of H2SO4 = 0.033L = 33.3 mL
Volume of H2SO4 = 33.3 mL


How many mL of 0.5M KMnO4 solution will react completely with 92.0g of K2C2O4.H2O
16H + 2MnO4+ + 5C2O42- → 10CO2 + 2Mn2+ + 8H2O

Solution
Moles of K2C2O4.H2O = 92/184 = 0.5 mol
5 mol of C2O42- = 2 mol of MnO4+
0.5 mol of C2O42- = 0.2 mol of MnO4+
Molarity of KMnO4 solution X Volume of KMnO4 solution = 0.2 mol of MnO4+
0.5 X Volume of KMnO4 solution = 0.2
Volume of KMnO4 solution = 0.4L = 400mL


What volume of 8.0M H2SO4 is needed to react completely with 108g of Al (Atomic Weight of Al = 27.0g)
2Al + 3H2SO4 → Al2(SO4)3 + H2

Solution
Moles of Al = 108/27 = 4mol
Number of moles of H2SO4 required for 4 mol Al = (3/2) X 4 = 6mol
Molarity X Volume of H2SO4 = Number of moles of H2SO4 required for 4 mol Al
8 X Volume of H2SO4 = 6
Volume of H2SO4 = 6/8 = 0.75L = 750mL


MolarityMolarity depends on temperature, dilution and volume
MolalityMolality depends only on dilution
Mole FractionMole fraction also depends only on dilution
NormalityNormality 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
2H2 + O2 → 2H2O

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 2H2 + O2 → 2H2O 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
N2 + 3H2 → 2NH3

Solution
1g of Nitrogen = 1/28 = 0.035 mol of Nitrogen
1g of Hydrogen = 1/2 = 0.5 mol of Hydrogen
From reaction equation N2 + 3H2 → 2NH3 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
CaCO3 → CaO + CO2

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 + 2H2 → CH4

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% H2SO4 by weight having density of 1.8g mL-1

Solution
Molarity = ((Volume Percentage) X 10 X (Density of H2SO4)) / Molecular weight of H2SO4
Molarity = 9.8 X 10 X 1.8 / 98 = 1.8M
Molarity = 1.8M


Find number of molecules in 15.8g of KMnO4

Solution
Number of moles in 15.8g of KMnO4 = 15.8 / 158 = 0.1 mol ( Molecular weight of KMnO4 = 158 )
Number of molecules = Number of moles X Avogadro Number
Number of molecules = 0.1 X 6.023 X 1023 = 6.023 X 1022
So, Number of molecules in 15.8g of KMnO4 = 6.023 X 1022


Calculate number of moles in 9.0g H2C2O4

Solution
Number of moles = Given Mass / Molecular Mass of H2C2O4
Molecular Mass of H2C2O4 = 1 X 2 + 12 X 2 + 16 X 4
Molecular Mass of H2C2O4 = 2 + 24 + 64 = 90
Molecular Mass of H2C2O4 = 90
Number of moles = 9/90 = 0.1 mol of H2C2O4


Calculate number of moles and number of molecules in 8.8g of CO2

Solution
Number of moles = Given Mass / Molecular Mass of CO2
Molecular Mass of CO2 = 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 1023
Number of molecules = 1.2046 X 1023


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

  • SO2 and SO3
  • NO2 and N2O
  • MgO and Mg(OH)2
  • NO and N2O5

Solution
SO2 and SO3 illustrates the law of multiple proportions here.


In a rocket motor fuelled with butane, 0.1 mol of butane requires what volume of O2 at STP for complete combustion.

Solution
Let’s see the reaction question of butane with oxygen
C4H10 + 13/2 O2 → 4CO2 + 5H2O
As per reaction equation 1 mol of C4H10 reacts with 13/2 mol of oxygen
So 0.1 mol of C4H10 reacts with 1.3/2 mol of oxygen
0.1 mol of C4H10 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
CaH2 + 2H2O → Ca(OH)2 + 2H2
2.1g of CaH2 would produce how much volume of H2 at STP.

Solution
As per Balanced Chemical Equation given above, 1 mol of CaH2 produces 2 mol of H2 meaning => 1 mol of CaH2 produces 2 X 22.4 litres of H2
1 mol of CaH2 produces 2 X 22.4 litres of H2
Let’s convert 2.1g of CaH2 to moles
Number of moles of CaH2 = 2.1 / 22 = 0.095
0.095 mol of CaH2 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:
P4 + 5O2 → P4O10
If 1.24g of P4 reacts with 8.0g of O2 then what amount of P4O10 is produced.

Solution
As per reaction 1 mol P4 reacts with 5 mol O2 meaning => 124g of P4 reacts with 160g of O2
So 1.24g of P4 reacts with 1.60g of O2 but 8.0g of O2 is given, that’s why for reaction Limiting Reagent would be P4
1 mol of P4 produces 1 mol of P4O10
124g of P4 produces 444g of P4O10
=> 1.24g of P4 produces 4.44g of P4O10
So amount of P4O10 produced in the reaction is 4.44g


Which of the following statements is/are wrong?
The following reactions occur:
CS2 + 3Cl2 → CCl4 + S2Cl2
1.0g of CS2 and 2.0g of Cl2 reacts with each other, then what amount of CCl4 is formed?

Solution
As per reaction equation 1 mol of CS2 reacts with 3 mol of Cl2
76g of CS2 reacts with 105g of Cl2
But as per Question, only we have only 1.0g of CS2 and 2.0g of Cl2
So 1.0g of CS2 reacts with 105/76 = 1.38g of Cl2 but we are given 2.0g of Cl2 . So for this reaction Limiting Reagent should be CS2
1 mol of CS2 produces 1 mol of CCl4
76g of CS2 produces 154g of CCl4
1.0g of CS2 produces 154/76 = 2.02g of CCl4
So amount of CCl4 produced in reaction is 2.02g


What amount of Al2O3 is produced if 108g of Al and 213g of MnO is reacted as per following reaction
2Al + 3MnO → Al2O3 + 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 Al2O3
Which means => 54g of Al produces 102g of Al2O3
108g of Al will produce (102 X 108)/54 = 204g of Al2O3
So in the reaction from 108g of Al and 213g of MnO only 204g of Al2O3 will be produced.


Which reactant is the limiting reagent in following reaction, if 21.0g of Li reacts with 32.0g of O2
4Li + O2 → 2Li2O

Solution
As per balanced reaction equation, 4 mol of Li reacts with 1 mol of O2
Meaning 28g of Li reacts with 32g of O2
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 O2
So 21.0g of Li actually required 24g of O2 to react with, but in the given reaction scenario 32.0g of O2 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 Cl2
2K + Cl2 → 2KCl

Solution
As per balanced chemical equation 2 mol of K reacts with 1 mol of Cl2
which means 39g of K reacts with 35g of Cl2
So, 3.9g of K will react with 3.5g of Cl2 , but as per question there’s 4.26g of Cl2 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 Na2CO3 reacts with 109.5g of HCl
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

Solution
As per balanced chemical equation above => 1 mol of Na2CO3 reacts with 2 mol of HCl
In other words, 108g of Na2CO3 reacts with 37.5g of HCl
So => 106g of Na2CO3 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 Na2CO3 would be Limiting Reagent.


27g of Al will react completely with ………………….g of O2 as per chemical equation 4Al + 3O2 → 2Al2O3

Solution
As per chemical equation => 4 mol of Al reacts with 3 mol of O2
108g of Al reacts with 96g of O2
So => 27g of Al will reacts with (96 X 27)/108 = 24g of O2


Calculate equivalent weight of H3PO4 in the following reaction
Ca(OH)2 + H3PO4 → CaHPO4 + 2H2O

Solution
As in the chemical reaction, only two of total three hydrogen atoms is replaced from H3PO4 . So equivalent weight of H3PO4 would be = Molecular Weight of H3PO4 / 2 = 98/2 = 49g
Equivalent Weight of H3PO4 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 250C, 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 BaCl2 is mixed with 0.20 mol of Na3PO4, then find out maximum number of moles of Ba3(PO4)2 which can be formed.

Solution
Balanced chemical equation of Reaction => 3BaCl2 + 2Na3PO4 → 6NaCl + Ba3(PO4)2
So 3 mol of BaCl2 reacts with 2 mol of Na3PO4
Given that we have only 0.5 mol of BaCl2 So Na3PO4 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 BaCl2 reacts with 0.33 mol of Na3PO4
But we are given only 0.20 mol of Na3PO4, that’s why in the reaction Na3PO4 will be Limiting Reagent

That’s why amount of Ba3(PO4)2 formed from reaction will depends upon Na3PO4 only.
As Per Balanced Chemical Equation => 2 mol of Na3PO4 produces 1 mol of Ba3(PO4)2
So 0.2 mol of Na3PO4 will produce 0.2 mol of Ba3(PO4)2
So => Maximum number of moles of Ba3(PO4)2 which can be produced from chemical reaction between BaCl2 and Na3PO4 is 0.2 mol.


If 0.5g of a mixture of two metals A and B displaces 560mL of H2 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 H2 = 22400mL = 2 Equivalents of H2
1 Equivalent of H2 = 11200mL
Number of H2 Equivalents in 560mL of H2 = 560/11200 = 0.05
Number of H2 Equivalents in 560mL of H2 = 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 H2
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 H2
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 Na2SO4.xH2O was found to contain 6.3g of H2O. The number of molecules of water of crystallisation is =>

  • 5
  • 7
  • 2
  • 10

Solution
Weight of Salt = 13.4g
Weight of H2O = 6.3g
Weight of anhydrous salt = 13.4 – 6.3 = 7.1g
Moles of anhydrous salt = 7.1/842 = 0.05
Moles of H2O = 6.3/18 = 0.35mol
0.05mol of anhydrous salt reacts with 0.35mol of H2O
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 H2SO4 is 18M and its density is 1.8g mL-1 then what will be molality of H2SO4?

Solution
density = (molarity of H2SO4 solution) X ((Molecular Weight of H2SO4 / 1000) + (1/molality of H2SO4))
1.8 = 18(98/1000 + 1/molality of H2SO4)
1.8 = 18(0.098 + 1/molality of H2SO4)
0.1 = 0.098 + 1/molality of H2SO4
0.002 = 1/molality of H2SO4
molality of H2SO4 = 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 H3PO4 = 98
Chemical Equation of Reaction between Magnesium Hydroxide and Phosphoric Acid
3Mg(OH)2 + 2H3PO4 → Mg3(PO4)2 + 3H2O
As Per Chemical Equation 3mol of Mg(OH)2 required for neutralising 2mol of H3PO4 .
So technically 1mol of Mg(OH)2 can be neutralised with 2/3 mol of H3PO4.
So 1mol = 58.3g of Mg(OH)2 can be neutralised by 2/3 mol = (2/3) X 98 = 65.33g of H3PO4
=> 58.3g of Mg(OH)2 can be neutralised by 65.33g of H3PO4
100g of Mg(OH)2 can be neutralised by (65.33/58.3) X 100 = 112.05g of H3PO4
So, for neutralising 100g of Magnesium Hydroxide 112.05g of Phosphoric acid is needed.


Recent Posts