
Q1 Calculate the molecular mass of the following: (i) H2O (ii) CO2 (iii) CH4 Ans: Molecules are made up of atoms & are quite small, therefore the actual mass of a molecule cannot be detrmined.It is expressed as the relative mass.C12 isotope is used to express the relative molecular masses of substances.Thus molecular mass of a substance may be defined as: the average relative mass of its molecule as compared to the mass of carbon atom taken as 12 amu.
(i) H_{2}O:
The molecular mass of water, H_{2}O
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 u
(ii) CO_{2}:
The molecular mass of carbon dioxide, CO_{2}
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 u
(iii) CH_{4}:
The molecular mass of methane, CH_{4}
= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u
Q2 Calculate the mass percent of different elements present in sodium sulphate (Na2SO4). Ans: The molecular formula of sodium sulphate is (Na_{2}SO_{4}).
Molar mass = atomic mass of Na(23) + atomic mass of S(32.066) + atomic mass of O(16)
Molar mass of (Na_{2}SO_{4}) = [(2 × 23.0) + (32.066) + 4 (16.00)] = 142.066
Mass percent of an element
∴ Mass percent of sodium:
Mass percent of sulphur:
Mass percent of oxygen:
Q3 Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. Ans: Empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound.
Form the available data Percentage of iron = 69.9
Percentage of oxygen= 30.1 Total percentage of iron & oxygen= 69.9+30.1= 100%
Step 1 calculation of simplest whole number ratios of the elements
Element
Percentage
Atomic mass
Atomic ratio
Simplest ratio
Simplest whole no ratio
Fe
69.9
55.84
69.9/55.84=1.25
1.25= 1
2
O
30.1
16
30.1/16 = 1.88
1.88=1.5
3
Step 2 writing the empirical formula of the compound
The empirical formula of the compound = Fe_{2} O_{3}
Q4 Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen. Ans: The chemical equation for the reaction is:
i) when 1 mole of carbon is burnt in 1 mole(32 g) of air(dioxygen) ,then according to chemical equation (1) 1mole(44 g) of carbon dioxide is produced.
ii) 32g of Oxygen upon heating gives CO2 = 44g
1 g of Oxygen upon heating gives CO2 = 44/32
Therefore 16 g of Oxygen upon heating gives CO2 = 44/32*16 = 22 g
(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.
Q5 Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1 Ans: 0.375 M solution of (CH_{3}COONa) = 0.375 moles(grams) of (CH_{3}COONa) dissolved in 1000 ml of solvent.
But according to question ,we have to make a 500 ml solution of (CH_{3}COONa)
∴Number of moles of sodium acetate in 500 mL
Molar mass of sodium acetate = 82.0245 g mole^{–1} (Given)
∴ Required mass of sodium acetate = (82.0245 g mol^{–1}) (0.1875 mole)
= 15.38 g
Q6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. Ans: Mass of nitric acid solution = 69 g (because 69% means 69 g dissolved in 100 ml of solvent)
Density of nitric acid solution = 1.41 g/ml
Volume of nitric acid solution = mass/density= 100/1.41 = 70.92 ml
No of moles of nitric acid = mass of nitric acid/molar mass of nitric acid=69/63= 1.09
Volume of the solution = 70.92
Molarity of the solution = moles of nitric acid/volume of solution in litres
= 1.09*1000/70.92= 15.37 moles/liter
∴Concentration of nitric acid = 15.37 mol/L
Q7 How much copper can be obtained from 100 g of copper sulphate (CuSO4)? Ans: 1 mole of CuSO_{4} contains 1 mole of copper.
Molar mass of CuSO_{4} = atomic mass of Cu + atomic mass of S + 4X atomic mass of O
Molar mass of CuSO_{4} = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 amu
Gram molecular mass of CuSO4= 159.5 g
159.5 g of CuSO_{4} contains Cu = 63.5
1 g of CuSO_{4} contains Cu= 63.5/159.5* 1
Therefore 100g of CuSO_{4} contains Cu = 63.5/159.5*100 = 39.81 g
Q8 Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol–1. Ans: From the available data Percentage of iron = 69.9
Percentage of oxygen= 30.1
Total percentage of iron & oxygen= 69.9+30.1= 100%
Step 1 calculation of simplest whole number ratios of the elements
Element
Percentage
Atomic mass
Atomic ratio
Simplest ratio
Simplest whole no ratio
Fe
69.9
55.84
69.9/55.84=1.25
1.25= 1
2
O
30.1
16
30.1/16 = 1.88
1.88=1.5
3
Step 2 Writing the empirical formula of the compound
The empirical formula of the compound = Fe_{2} O_{3}
Step 3 determination of molecular formula of the compound
Empirical formula mass = 2 X69.9 + 3 X16=187.8 amu
Molecular mass of oxide= 159.69g/mol(given)
Now we know molecular formula = n x Empirical formula
And n= molecular mass / empirical formula mass= 159.69/187.8 = 0.85 = approx 1
Therefore molecular formula = n x empirical formula
=1 x(Fe_{2}O_{3}) = Fe_{2}O_{3}
The molecular formula of the oxide is Fe_{2}O_{3}
Q9 Calculate the atomic mass (average) of chlorine using the following data: % Natural Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659 Ans: Atomic mass of first isotope = 34.9689
Natural abundance of first isotope = 75.77% or 0.757
Atomic mass of second isotope= 36.9659
Natural abundance of second isotope= 24.23% or 0.242
Now average atomic mass of chlorine
= [34.9689x 0.757 + 36.9659x 0.242]/(0.757+0.242) =35.4521
So, the average atomic mass of chlorine = 35.4527 u
Q10 In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane. Ans: Mole is a latin term which means a pile or mass of stones placed in sea. In Chemistry a mole represents Avogadro’s number(6.022 X 10^{23} particles)
(i) 1 mole of C_{2}H_{6} contains 2 moles of carbon atoms(because 2 carbons are present in the formula of ethane)
∴ Number of moles of carbon atoms in 3 moles of C_{2}H_{6}
= 2 × 3 = 6
(ii) 1 mole of C_{2}H_{6 }contains 6 moles of hydrogen atoms(because 6 hydrogen atoms are present in ethane)
∴ Number of moles of carbon atoms in 3 moles of C_{2}H_{6}
= 3 × 6 = 18
(iii) 1 mole of C_{2}H_{6 }contains 6.023 × 10^{23} molecules of ethane(because 1mole = 6.022 x 10^{23})
∴ Number of molecules in 3 moles of C_{2}H_{6}
= 3 × 6.023 × 10^{23} = 18.069 × 10^{23 }
Q11 What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2 L? Ans: Mass of sugar= 20g
Molecular mass of sugar=12x12 +1x22 +16x11=342 g
Mass of water= 2 liter
Molarity (M) = number of moles of solute/volume of solution in liters
= 20/(342x2) =0.02924 mol/liter
Therefore Molar concentration of sugar = 0.03 mol L^{–1}
Q12 If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution? Ans: Let the volume of 0.25 M of ethanol needed = x Molar mass of methanol (CH^{3}OH) = (1 × 12) + (4 × 1) + (1 × 16) = 32 g mol^{–1} = 0.032 kg mol^{–1}
Molarity of methanol solution
= 24.78 mol L^{–1} (Since density is mass per unit volume)
Applying, M1V1 = M2V2
(Given solution) (Solution to be prepared)
(24.78 mol L^{–1}) x = (2.5 L) (0.25 mol L^{–1})
x = 0.0252 L
x = 25.22 mL
Q13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below: 1Pa = 1N m–2 If mass of air at sea level is 1034 g cm–2, calculate the pressure in Pascal. Ans: Pressure is defined as force acting per unit area of the surface.
= 1.01332 × 10^{5} kg m^{–1}s^{–2}
We know,
1 N = 1 kg ms^{–2 } Then,
1 Pa = 1 Nm^{–2} = 1 kg m^{–2}s^{–2}
1 Pa = 1 kg m^{–1}s^{–2}
1 Pressure = 1.01332 × 10^{5} Pa
Q14 What is the SI unit of mass? How is it defined? Ans: The system of measurement is the most common system employed all over the world.The 11^{th} General Conference of the Weights & Measures held in France in 1960 introduced this system called Systeme International d’ units, abbreviated as SI system.
The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.
Q15 Prefixes Multiples (i) micro 106 (ii) deca 109 (iii) mega 10–6 (iv) giga 10–15 (v) femto 10 Ans: SNO
PREFIXES
SYMBOLS
MULTIPLES
1
Micro
μ
10^{6}
2
Deca(deka)
da
10^{1}
3
Mega
M
10^{6}
4
Giga
G
10^{9}
5
Femto
f
10^{15}
Q16 What do you mean by significant figures? Ans: Significant figures are those meaningful digits that are known with certainty.
The significant figures in any number are all certain digits plus one doubtful digit. In order to report a scientific data, the data is expressed in significant figures in which all the digits reported are certain except the last one which is uncertain or doubtful.For eg 20.86 has four digits In all.Out of them 2,0 & 8 are certain digits while the last digit 6 is uncertain.Thus the number 20.86 has all the four digits as significant figures.Out of them 2, 0 & 8 are certain while 6 has some uncertainity about it.
Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result.
Q17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass. (ii) Determine the molality of chloroform in the water sample. Ans: (i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.
∴ Mass percent of 15 ppm chloroform in water
(ii) molality (M) = no of moles of solute/mass of solvent in g *1000
Therefore mass of chloroform= 12 + 1+3(35.5) = 119.5 g/mol
100 g of the sample contains 1.5 × 10^{–3} g of CHCl_{3.}
⇒ 1000 g of the sample contains 1.5 × 10^{–2} g of CHCl_{3.}
m = 1.5 x 10^{3}/119.5 * 1000 = 1.25x 10^{4} m
Q18 Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012 Ans: In scientific notation every number is written as N x 10^{n}, where N = a number with single non zero digit to the left of the decimal point n = exponent of 10, which may be positive, negative integer or zero
(i) 0.0048 = 4.8× 10^{–3}
(ii) 234, 000 = 2.34 ×10^{5}
(iii) 8008 = 8.008 ×10^{3}
(iv) 500.0 = 5.000 × 10^{2}
(v) 6.0012 = 6.0012
Q19 How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034 Ans: The significant figures in any number are all certain digits plus one doubtful digit.Certain rules are followed which are as under:
1 all non zero digits in anumber are significant
2 the zeros between two non zero digits are always significant
3 the zeros written to the left of the first non zero digit in a number are non significant.They simply indicate the position of the decimal point.
4 all zeros placed to the right of a decimal point in a number are significant.
(i) 0.0025
There are 2 significant figures.
(ii) 208
There are 3 significant figures.
(iii) 5005
There are 4 significant figures.
(iv) 126,000
There are 3 significant figures.
(v) 500.0
There are 4 significant figures.
(vi) 2.0034
There are 5 significant figures.
Q20 Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808 Ans: Rounding off a number means that the digits which are not significant have to be dropped.The rules are as follows:
 if the digit to be dropped is more than 5, then add 1 to the preceding significant figure.For eg in the number 11.06, the digit to be dropped is 6.Therefore the preceding digit 0 is increased by one & the final result is 11.1.
 if the digit to be dropped is less than 5,then it is deleted as such without bringing any change in the preceding significant figure.For eg 43.123 is to reported upto four significant figures,then the last digit 3 is dropped & the final answer is 43.12.
 if the digit to be dropped is 5 then the preceding significant digit in the number may be left unchanged if it is even & may be increased by 1 in case if it is odd.Eg 1.6145, if reported upto four significant figures will become 1.614 after rounding off. Similarly 1.6175 will become 1.618.
 if during rounding off, more than one digit is to be dropped form a particular number, then they are dropped on at a time by following the above rules.
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810
Q21 The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g (a) Which law of chemical combination is obeyed by the above experimental data?Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ...................... mm = ...................... pm (ii) 1 mg = ...................... kg = ...................... ng (iii) 1 mL = ...................... L = ...................... dm3 Ans: Let us fix 14 parts by weight of nitrogen as fixed weight.
Now let us calculate the weights of oxygen which combine with 14 parts by weight of nitrogen
Sno
No of parts by weight of nitrogen
No of parts by weight of oxygen
14 parts of nitrogen as fixed weight
No of parts by weight of oxygen which combine with 14 parts by weight of nitrogen
1
14g
16g
14g
16
2
14g
32g
14g
32
3
28g
32g
14g
32
4
28g
80g
14g
80
(a) If we fix the mass of dinitrogen at 14 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 16 g, 32 g, 32 g, and 80 g.
The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions.
This law was given by Dalton in 1804. The law states that if two elements combine to form 2 or more compound, then the weight of one element which combines a fixed weight of other element in these compounds,bears a simple whole number ratio by weight.
(b) (i) We know 1km=1000m
Or 1m = 1000 mm
Therefore 1km = 1000x 1000mm= 10^{6} mm
1 km = 1 km ×
1 km = 10^{15} pm
Hence, 1 km = 10^{6} mm = 10^{15} pm
(ii) We know 1kg = 1000mg
Or 1000mg= 1kg
Or 1mg= 1/1000* 1= 0.01 kg
1 mg = 1 mg ×
⇒ 1 mg = 10^{6} ng
1 mg = 10^{–6} kg = 10^{6} ng
(iii) We know 1000 ml=l L
Or 1ml=1/1000*1= 0.01L
1 mL = 1 cm^{3} = 1 cm^{3}
⇒ 1 mL = 10^{–3} dm^{3}
1 mL = 10^{–3} L = 10^{–3} dm^{3}
Q22 If the speed of light is 3.0 × 108 ms–1, calculate the distance covered by light in 2.00 ns. Ans: According to the question:
Time taken to cover the distance = 2.00 ns
= 2.00 × 10^{–9} s
Speed of light = 3.0 × 10^{8} ms^{–1}
We know speed = distance/time
Or distance = speed x time
Therefore Distance travelled by light in 2.00 ns
= Speed of light × Time taken
= (3.0 × 10^{8} ms ^{–1}) (2.00 × 10^{–9} s)
= 6.00 × 10^{–1} m
= 0.600 m
Q23 In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B (iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B (v) 2.5 mol A + 5 mol B Ans: while solving the problems relating to chemical equation,the reactant react according to the balanced chemical equation.Quite often, one of the reactant is present in lesser amount while the other may be present in higher amount.The reactant which is present in lesser amount is known as limiting reactant or reagent.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.Here atom B is in lesser amount(200)
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent because B is less as compared to A
(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.
Q24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2(g) + H2(g) → 2NH3(g) (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass? Ans: (i) Balancing the given chemical equation,
Total mass of Ammonia = 2((14) +3(1))= 34 g
From the chemical equation,we can write
28gm of N_{2} reacts with 6gm of H_{2} to produce ammonia= 34g
Or 1 gm of N_{2} reacts with 1gm of H_{2} to produce ammonia= 34/28*1
Or when 2.00x10^{3} g of N_{2} reacts with 1.00x10^{3} gm of H_{2} to produce ammonia
=34/28 *2.00x10^{3}=2428.57g
hence 2.00 × 10^{3} g of dinitrogen will react with 1.00x10^{3} g of dihydrogen to give 2428.57 g of ammonia
Given, Amount of dihydrogen = 1.00 × 10^{3} g
Hence, N_{2} is the limiting reagent.
(ii) N_{2} is the limiting reagent and H_{2} is the excess reagent. Hence, H_{2} will remain unreacted.
(iii) Mass of dihydrogen left unreacted = 1.00 × 10^{3} g – 428.6 g
= 571.4 g
Q25 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different? Ans: Molar mass of Na_{2}CO_{3} = (2 × 23) + 12.00 + (3× 16) = 106 g mol^{–1}
1 mole of Na_{2}CO_{3} = 106 g
0.5 mol of Na_{2}CO_{3 }= 53 g Na_{2}CO_{3}
⇒ 0.50 M of Na_{2}CO_{3} = 0.50 mol/L Na_{2}CO_{3}
Hence, 0.50 mol of Na_{2}CO_{3} is present in 1 L of water or 53 g of Na_{2}CO_{3} is present in 1 L of water.
Q26 If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced? Ans: Reaction of dihydrogen with dioxygen can be written as:
6H_{2} + 3O_{2} = 6H_{2}O
Now, SIX volumes of dihydrogen react with THREE volume of dihydrogen to produce SIX volumes of water vapour.
Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.
Q27 Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg Ans: (i) 28.7 pm:
1 pm = 10–12 m
28.7 pm = 28.7 × 10–12 m
= 2.87 × 10–11 m
(ii) 15.15 pm:
1 pm = 10–12 m
15.15 pm = 15.15 × 10–12 m
= 1.515 × 10–12 m
(iii) 25365 mg:
1 mg = 0.01 kg
Therefore 25365 mg = 0.01/1 x 25365= 253.65 kg
Q28 Which one of the following will have largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g) Ans: (i) Gram atomic mass of Au= 197 g
Or
197g of Au contains = 6.022 x 10^{23}
Therefore 1gm of Au contains = 6.022 x 10^{23}/197*1 = 3.06 x 10^{21} atoms
(ii) Gram atomic mass of Na = 23 g
Or
23 g of Na contains atoms = 6.022x 10^{23}
Or
1gm of Na contains atoms = 6.022x10^{23} /23 *1 = 26.2 x10^{21} atoms
(iii) Gram atomic mass of Li = 7
Or
7g of Li contains atoms = 6.022 x 10^{23}
Or
1g of Li contains atoms = 6.022 x 10^{23}/7 *1= 86.0 x 10^{21} atoms
(iv) Gram atomic mass of Cl = 71 Or 71g of Cl contains atoms = 6.022x10^{23}
Or
1 g of Cl contains atoms = 6.022x10^{23} /71 * 1= 8.48 x 10^{21} atoms
Hence, 1 g of Li (s) will have the largest number of atoms.
Q29 Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). Ans: Mole fraction of C_{2}H_{5}OH= .................1
Let the moles of C_{2}H_{5}OH= X
Now density of water = 1 (given)
And the weight of 1000ml of water = volume * density (from density = mass/volume)
= 1000 x 1=1000g
Therefore moles of water = 1000/18= 55.55 mol (18g is molecular mass of water)
Also mole fraction of C_{2}H_{5}OH= 0.040 (given)
Putting the values in equation 1,we get
=0.040=X/X+55.55
=0.040X+2.222 = X
OR X= 2.3145 mol
Molarity of solution = 2.314 M
Q30 What will be the mass of one 12C atom in g? Ans: 1 mole of carbon= 6.022x 10^{23} carbon atoms = 12 g carbon
Or 6.022x 10^{23} atoms of carbon = 12 g of carbon
Therefore 1 atoms of carbon = 12/6.022x 10^{23} x 1= 1.993 x 10^{23} g of carbon
Q31 How many significant figures should be present in the answer of the following calculations? (i) (ii) 5 × 5.364 (iii) 0.0125 + 0.7864 + 0.0215 Ans: Rules for determining the significant figures
In addition or subtraction of the numbers having different precisions, the final result should be reported to the same number of decimal places as in the term having least number of decimal places
(i) = 1.648
Least precise number of calculation = 298.15
Number of significant figures in the answer
= Number of significant figures in the least precise number = 2 i.e 1.65
(ii) 5 × 5.364 = 26.820
Least precise number of calculation = 5.364
Number of significant figures in the answer = Number of significant figures in 5.364 = 3 i.e 26.820
(iii) 0.0125 + 0.7864 + 0.0215 = 0.8204
Since the least number of decimal places in each term is four, the number of significant figures in the answer is also 4 i.e 0.8204
Q32 Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755 gmol–1 0.337% 38Ar 37.96272 gmol–1 0.063% 40Ar 39.9624 gmol–1 99.600% Ans: Molar mass of argon =
Atomic mass of ^{36}Ar = 35.96755 & abundance = 0.337
Or
total atomic mass of ^{36}Ar = 35.96755 * 0.337= 0.121g/mol
Atomic mass of ^{38}Ar = 37.96272 & abundance = 0.063
Or
total atomic mass of ^{38}Ar = 37.96272 * 0.063= 0.024g/mol
Atomic mass of ^{40}Ar =39.9624 & abundance = 99.600
Or
total atomic mass of ^{40}Ar = 39.9624 * 99.600= 39.802g/mol
Therefore molar mass of argon = total mass of ^{36}Ar+ total mass of ^{38}Ar + atomic mass of ^{40}Ar
= 0.121 + 0.024 + 39.802= 39.947 g/mol
Q33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He. Ans: (i) 1 mole of Ar = 6.022 × 10^{23} atoms of Ar
52 mol of Ar = 52 × 6.022 × 10^{23} atoms of Ar
= 3.131 × 1025 atoms of Ar
(ii) Atomic mass of He = 4amu
Or
4amu is the mass of He atoms = 1
Therefore 52 amu is the mass of He atoms= ¼*52 = 13 atoms of He
(iii) Gram atomic mass of He = 4g
Or
4g of He contains = 6.022x 10^{23} atoms
Therefore 52 g of He contains = 6.022x 10^{23} /4 * 52 = 7.83 x 10^{24} atoms
Q34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula. Ans: (i) percentage of C can be calculated as follows:
CO_{2} = C
i.e 44 parts of CO_{2}= 12 parts of C
OR
44g of CO_{2} = 12 g of C
Therefore according to question
3.38 g of CO_{2} contains C = 12/44 * 3.38 = 0.921 g
18 g of water contains hydrogen = 2g
Therefore 0.690 g of water contains hydrogen = 2/18 * 0.690 = 0.0767g
0.690 g of water will contain hydrogen
= 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:
= 0.9217 g + 0.0767 g = 0.9984 g
Now percentage of carbon = weight of carbon/weight of compound * 100
=0.921 /0.998 * 100= 92.32 %
Also percentage of hydrogen = weight of hydrogen/weight of compound *100
=0.0766/0.998 * 100 = 7.68 %
(ii) Given,
Weight of 10.0L of the gas (at S.T.P) = 11.6 g
Weight of 22.4 L of gas at STP
= 25.984 g
≈ 26 g
Hence, the molar mass of the gas is 26 g.
(iii) empirical formula
Element Percentage Atomic mass Atomic ratio Simplest ratio Simplest whole no ratio C 92.32 12 92.32/12 = 7.69 7.69/7.65 = 1.00 1 H 7.65 1 7.65/1 = 7.65 7.65/7.65 = 1 1 Empirical formula of the compound = CH
Now molecular formula calculation
Empirical formula mass = 12 + 1 = 13 amu
Also molecular mass = 26 g (calculated in previous step)
Therefore n = molecular mass/empirical formula mass = 26/13 = 2
Now molecular formula = n x empirical formula = 2 x CH = C_{2}H_{2}
Q35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? Ans: 0.75 M of HCl ≡ 0.75 mol of HCl X molecular weight of HCl dissolved in 1000 ml of water
Or
[(0.75 mol) × (36.5 g mol^{–1})] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains Hcl = 27.375g
Or
1 ml of solutions contains Hcl = 27.375/1000 * 1
And 25 ml of solutions contains Hcl = 27.375/1000 * 25 = 0.6844 g.
From the given chemical equation,
CaCO_{3(s)} + 2 HCl_{(aq)} → CaCl_{2(aq)} + CO_{2(g)} + H_{2}O_{(l)}
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO_{3} (100 g).
Amount of CaCO_{3} that will react with 0.6844 g= 0.9639 g
Q36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g) How many grams of HCl react with 5.0 g of manganese dioxide? Ans: Molecular weight of Mno_{2} = 87 g
Molecular weight of Hcl = 4 * 36.5 = 146 g
According to chemical reaction
1 g of Mno_{2} reacts with 4g of Hcl togive = 146/87 * 1
Or
5 g of Mno_{2} reacts with 4g of Hcl to give = 146/87 *5 = 8.4 g of Hcl
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.