Rajasthan Board RBSE Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry Textbook Exercise Questions and Answers.
Rajasthan Board RBSE Solutions for Class 11 Chemistry in Hindi Medium & English Medium are part of RBSE Solutions for Class 11. Students can also read RBSE Class 11 Chemistry Important Questions for exam preparation. Students can also go through RBSE Class 11 Chemistry Notes to understand and remember the concepts easily.
Question 1.1.
Calculate the molecular mass of glucose (C_{6}H_{12}O_{6}) molecule.
SolutIon:
Atomic mass of C = 12.011 u
Atomic mass of H = 1.008 u
Atomic mass’of O = 16.00 u
Molecular mass of C_{6}H_{12}O_{6}
= 6 x (12.011 u) + 12 x (1.008 u) + 6 x (1600 u)
= 72.066 u + 12.096 u + 9600 u
= 180.162 u
Question 1.2.
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular fomula?
Solution:
∴ Empirical formula of compound = C_{1}H_{2}Cl_{1}
= CH_{2 }Cl
Empirical formula mass = 12 x 2 (1) + 35.5
= 49.5
Moleular mass = 98.96
= \(\frac{98.96}{49.5}\) = 2
Molecular formula = n x Empirical formula
= 2 x CH_{2}Cl
= C_{2}H_{4}Cl_{2}
Question 1.3.
Calculate the amount of water (g) produced by the combustion of 16 g of methane.
Solution:
The balanced equation for the combustion of
methane (CH_{4}) is
CH_{4}(g) + 2O_{2}(g) → CO_{2}(g) + 2H_{2}O(g)
16g of CH_{4} = lmole of CH_{4}
∵1 mol of CH_{4}(g) gives 2 mol of H_{2}O(g).
2 mol of water(H_{2}O) = 2 x (+16)
= 2 x 18 = 36g
1 mol of H_{2}O 18g H_{2}Os = \(\frac{18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}\) = 1
∴ 2 moles H_{2}O x \(\frac{18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}\) = 2 x 18gH_{2}O = 36gH_{2}O
Question 14.
How many moles of methane are required to produce 22 g CO_{2} after combustion?
Solution:
The chemical reection for combustion of methane is:
∵ 44g CO_{2} (g) is obtained from 16g CH_{4}(g)
[1 mol CO_{2}(g) is obtained from 1 mol of CH_{4}(g)]
Number of moles of CO_{2}(g)
= 22gCO_{2}(g) x \(\frac{1 \mathrm{molCO}_2(g)}{44 \mathrm{~g} \mathrm{CO}_2(g)}\)
= 0.5 mol CO_{2}(g)
∴ 0.5 mol CO_{2} (g) would be obtained from 0.5 mol CH_{4}(g)
Hence, 0.5 moI of meihane is required to porduce 22g of CO_{2}(g) after combustion.
Question 1.5.
50.0 kg of N_{2}(g) and 10.0 kg of H_{2}(g) are mixed to produce NH_{3}(g). Celculate the amount of NH_{3} formed. Identify the limiting reagent in the production of NH_{3} in this reaction.
Solution:
A balanced equetion is
Number of moles of N_{2}
= 5OkgN_{2} x \(\frac{1000 \mathrm{~g} \mathrm{~N}_2}{1 \mathrm{~kg} \mathrm{~N} \mathrm{~N}_2} \times \frac{1 \mathrm{~mol} \mathrm{~N} \mathrm{~N}_2}{28 \mathrm{~g} \mathrm{~N}_2}\)
= 1786 x 10^{2} mol
Number of moles of H_{2}
= 10kgH_{2} x \(\frac{1000 \mathrm{~g} \mathrm{H}_2}{1 \mathrm{~kg} \mathrm{H}_2} \times \frac{1 \mathrm{~mol} \mathrm{H}_2}{2.016 \mathrm{~g} \mathrm{H}_2}\)
= 496 x 10^{3} mol
∵1 mol N_{2}(g) requires 3 moles H_{2}(g) for the reaction
∴ For 17.86 x 10^{2} mol of N_{2},
No. of moles of H_{2}(g)
= 1786 x 10^{2}molN_{2} x im
= 536 x 10^{3 }molH_{2}
But we have only 4.96 x 10^{3} mol H_{2}. Therefore. dihydrogen is the limiting reagent in this case. Therefore, NH_{3}(g) would be obtained only from that
amount of available dihydrogen i.e. 4.6 x 10^{3} mol
∵ 3 mol H_{2} (g) gives 2 mol NH_{3} (g)
∴ 4.96 x 10^{3} mol H_{2}(g) will give
= 4.96 x 10^{3} mol H_{2}(g) x \(\frac{3 \mathrm{~mol} \mathrm{H}_2(g)}{1 \mathrm{~mol} \mathrm{~N}_2(g)}\)
= 3.30 x 10^{3} mol NH_{3}(g)
lmolNH_{3}(g) = 17.OgNH_{3}(g)
3.30 x 10 mol NH_{3}(g) x \(\frac{17.0 \mathrm{~g} \mathrm{NH}_3(g)}{1 \mathrm{molNH}_3(g)}\)
= 3.30 x 10^{3} x 17g NH_{3}(g)
= 56.1 x 10^{3}g NH_{3}(.g)
= 56.1 kg NH_{3}(g)
Question 1.6.
A solution is prepared by adding 2 g of a substance ‘A’ to 18 g of water, calculate the mass percent of the solute.
SolutIon:
Mass of substance ‘A’ =2 g
Mass of water = 18g
Question 1.7.
Calculate the molarity of NaOH in the solution prepared by dissolving its 4g in enough water to form 250 mL of the solution.
SolutIon:
Mass of NaOH = 4g
Molar mass of NaOH = 23 + 16 +1 =40
Volume of solution = 250 mL
Molarity = ?
Question 1.8.
The density of 3 M solution of NaCl is 1.25 g mL^{1}. Calculete the molality of the solution.
Solution:
M = 3 mol L’
Molar mass of NaCI = 23 + 35.5 = 58.5
Mas8 of NaC1 in iL solution = 3 x 58.5 = 175.5 g
Mass of 1 L solution = 1000 x 125 = 1250 g
(∵Mass = volume x density)
Mass of water in solution = 1250 - 175.5
= 10745 g
= 279 m
Question 1.1.
Calculate the molar mass of the following:
(i) H_{2}O
(ii) CO_{2}
(iii) CH_{4}
SolutIon:
(i) Molar mass of H_{2}O
= 2 x (1.008u) + 16.00 u
= 2.016 u + 16.00 u
= 18.02 u
(ii) Molar mass of CO_{2}
= 12.011 u + 2 x (16.00 u)
= 12.011 u + 32.00 u
= 44.O1u
(iii) Molar mass of CH_{4}
=12.011 u + 4 x (1.008 u)
= 12.011 u + 4.032 u
= 16.0434 u
Question 1.2.
Calculate the mass percent of different elements present n sodium sulphate (Na_{2}SO_{4}).
Solution:
Molar mass of Na_{2}SO_{4}
= 2 x (23.00 u) + 32.066 u + 4 x (16.00 u)
= 46.00 u + 32.066 u + 6400u
= 142.066 u
Question 1.3.
Determine the empirical formula of an oxide of iron, vhich has 96.9% iron and 30.1% dioxygen mass.
Solution:
∴ Empirical formula of oxide of iron.
= Fe_{2}O_{3}.
Question 1.4.
Calculate the amount of carbon dioxide that could be produced when:
(1) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16g of dioxygen.
(iii) 2 molar of carbon are burnt in 16 g of dioxygen.
SolutIon:
Balanced equation is:
(i) From the above reaction
1 mol of carbon on burning in air produces
1 mol of CO_{2} i.e. 44g of CO_{2}
(ii) When 1 mo1 of carbon is burnt in 16 g of dioxygen then 22g of CO will be produced. In this case, dioxygen
is a limiting reagent.
∵32 g of O_{2} gives 44 g of CO_{2}
∵ 16 g of O_{2} will give = \(\frac{44}{32} \times 16\)
= 22 g of CO_{2}
(iii) As dioxygen is a limiting reagent, therefore, 16 g of dioxygen can react only with 0.5 mol of carbon to produce
22 of CO_{2}.
Question 1.5.
Calculete the mass of sodium acetate (CH_{3}COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mo1^{-1}.
Solution:
Volume of solution 500 mL
Molarity = 0.375 M
Molar mass of sodium acetate (M) = 82.0245 g mol^{-1}
= 15.379 g
Question 1.6.
Calculate the concentration of nitric acid in moles per litre in a sample which has a density l.41g mU’ and the mass percent of nitric acid in it being 69%.
Solution:
Density of solution = 1.41 g mL^{-1}
Mass % of HNO_{3} = 69%
69% of HNO_{3} means loo g of its solution contains 69 g HNO_{3}
∴ Mass of solution =100 g
Mass (w) of HNO_{3} = 69 g
Moler mass (M) of HNO_{3} = L008 + 14.007 + 3 x 16.00
= 63.015g mol^{-1}
= 15.439 M
Question 1.7.
How much copper can be obtained from loo g of copper sulphate (CuSO_{4})?
Solution:
Mass of CuSO_{4} 100 g
Molar mass of CuSO_{4} = 63.5 + 32 + 4 x 16
= 159.5g mol^{-1}
1 mol of CuSO_{4 }contains 1 mol of Cu.
∵159.5 g of CuSO_{4} forms = 63.5 g of Cu.
∵ 100 g of CuSO_{4} will form = \(\frac{63.5 \mathrm{~g}}{159.5 \mathrm{~g}}\)
= 39.81 g of Cu.
Question 1.8.
Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
Solution:
∴ Empirical formula of oxide of iron = Fe_{2}O_{3}
Empirical formula mass = 2 x 55.85 + 3 x 16.00
= 159.1 g mol^{-1}
\(=\frac{159.8}{159.7}=1\)
Molecular formula = n x Empirical formula
= n x Fe_{2}O_{3}
= 1 x Fe_{2}O_{3}
= Fe_{2}O_{3}
Question 1.9.
Calculate the atomið mass (average) of chlorine using the following data:
%Natural |
Abundence |
Molar mass |
35Cl |
75.77 |
34.9689 |
35Cl |
24.33 |
36.9659 |
Solution:
Fractional abundance òf ^{35}Cl
\(=\frac{75.77}{100}\) = 0.7577
Molar mass of ^{35}Cl = 34.9689
Molar mass Cl = 36.9659
∵ Average atomic mass of chlorine
= Fractional abundance òf ^{35}Cl x Molar mass of ^{35}C1 +
Fracttonal abundance of Cl x Molar mass Cl
= 0.7 577 x 34.9689 + 02423 x 36.9659
= 26.4959 + 8.9568
= 35.4527 amu
Question 1.10.
In three moles of ethane (C_{2}H_{6}), calculate the following:
(i) Number of moles of carbon atoms
(ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane
Solution:
(i) ∵ 1 mol of ethane (C_{2 }H_{6} )contain 2 mol of carbon atoms.
∴ 3 mol of C_{2}H_{6} will contain = 2 x 3 = 6 mol of carbon atoms
(ii) ∵ 1 mol of ethane (C_{2 }H_{6} )contains 6 mol of hydrogen atoms.
∴ 3 mol of C_{2}H_{6} will contain = 6 x 3 = 18 mol of hydrogen atons.
(iii) ∵ 1 mol of ethane (C_{2} H_{6}) contains 6.022 x 10^{23} molecules of ethane.
∴3 moles of C_{2}H_{6} will contain
= 3 x 6.022 x 10^{23}
= 18.066 x 10^{23} molecules
= 1.8066 x 10 molecules
Question 1.11.
What is the concentration of sugar (C_{12}H_{22}O_{11}) in mol L^{-1} if its 20 g are dissolved in enough water to make a final volume up to 2L?
Solution:
Mass of C_{12}H_{22}O_{11}(w) = 20 g
Volume of solution = 2L
Molar mass ofC_{12}H_{22}O_{11}
M = 12 x 12.01 + 22 x 1.008 x 11 x 16.0O
= 34229
= 0.029 M
Question 1.12.
f the density of methanol is 0.793 kg L’, what is its volume needed for making 2.5 L
of its 0.25 M solution?
Solution: Denstiy of methanol = 0.793 kg L’
= 0.793 X 10^{3}g L^{-1}
Molar mass of methanol (CH_{3}OH)
= 1 x 12.01 + 4 x 1.008 + 16.00
= 32.042
= 24.749 molL^{1}
∴ Initial molarity (M_{1}) = 24.749 M
Initial volume (V_{1}) = ?
Final molarity (M_{2}) = 025 M
Final volume (V_{2}) = 2.5 L
∵ M_{1}V_{1} = M_{2}V_{2}
24.749 x V_{1} = 0.25 x 2.5
\(V_1=\frac{0.25 \times 2.5}{24.749}\)
= 25.25 ML
Question 1.13.
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa = lNm^{-2}
If mass of air at sea level is 1034 g cm^{2}, calculate the pressure in pascal.
Solution:
Mass of air at sea level =1034 g cm^{-2}
As we know that,
Acceleration due to gravity
g = 9.8 ms^{-2}
Pressure of air = \(\frac{1034}{1000}\) x 9.8 x 100 x 100
= 1.012322 x 10^{5} pa
Question 1.14.
What is the SI unit of mass? How is it defined?
Solution:
The S.J unit of mass is kilogram (kg). It is defined as the amount of matter present in a substance.
Question 1.15.
Match the following prefixes with their multiples:
Prefixes |
Multiples |
Micro |
10^{6} |
deca |
10^{9} |
mega |
10^{-6} |
giga |
10^{-15} |
femio |
10 |
Solution:
Prefixes |
Multiples |
Micro |
10^{6} |
deca |
10^{9} |
mega |
10^{-6} |
giga |
10^{-15} |
femio |
10 |
Question 1.16
What do you mean by significant figures?
Solution:
The total number of digits in a number, which are meaningful with certainty are called significant figures. e.g 3.016 has four significant figures.
Question 1.17.
A sample of drinking water was found to be severely contaminated with chloroform, CHCl_{3}, 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.
Solution:
(i) 15 ppm means 15 parts in 10^{6} parts
Mass % = \(\frac{15 \times 100}{10^6}\) = 1.5 x 10^{3}%
(ii) Molar mass of CHCl_{3} (M)
= 12.01 + 1.008 + 3 x 35.5
= 119.36gmol^{-1}
1.5 x 10% means 1.5 x 10^{3}g CHCl_{3} is found in loo g sample.
= 1.26 x 10^{4}m
Question 1.18.
Express the following in the scientific notation:
1. 0.0048
2. 234,000
3. 80088
4. 500.0
5. 6.0012
Solution:
Question 1.19.
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 |
Solution:
Number |
Significant figures |
(i) 0.0025 |
2 |
(ii) 208 |
3 |
(iii) 5005 |
4 |
(iv) 126,000 |
3 |
(v) 500.0 |
4 |
(vi) 2.0034 |
5 |
Question 1.20.
Round up the following upto three significant figures:
1. 34.216
2. 10.4107
3. 0.04597
4. 2808
Solution:
Question 1.21.
The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dioxygen |
Mass of dinitrogen |
(i) 14 g |
16g |
(ii) 14 g |
32g |
(iii) 28 g |
32g |
(iv) 28 g |
80g |
(a) Which law of chemical combination is obeyed by the above experimental data?
(b) Fill in the blanks in the following conversions:
(i) 1 km = ................ mm = ............... pm
(ii) 1mg = ................ kg ............... ng
(iii) 1 mL.=................ L = ................ dm3
Solution:
Mass of dioxygen |
Mass of dinitrogen |
(i) 14 g |
16g |
(ii) 14 g |
32g |
(iii) 28 g |
32g |
(iv) 28 g |
80g |
The above experimental data obey law of multiple proprotions.
According to this law," when two elements combine to form more than one compound, then the masses of one of the elements which combine with fixed mass of other element are in the simple whole number ratio,"
(b) (i) 1 km = 10 mm = 10^{15} pm
(ii) 1 mg = 10 kg = 10^{6} ng
(iii) 1 ml = 10^{-3}L= 10^{-3} dm^{3}
Question 1.22.
If the speed of light is 3.0 x 10 ms1, calculate the distance covered by light in 2.00 ns.
Solution:
Speed of light = 3.0 x 10 ms^{-1}
Time = 2.00 ns
= 2 × 10 9 8
∵ Distance = speed x time
= 30 x 10 ms 1 x 2 x 10^{-9} s
= 60 × 10^{-1} m
= 0.60 m
Question 1.23.
In a reaction,
A + B_{2} → AB_{2}
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
Solution:
A + B_{2} → AB_{2}
(i) In above rection,
∵1 atom of A reacts with 1 molecule of B.
∴200 atoms of A will react with 200 moleculers of B. Hence, in this case, 'B' is the limiting reagent, as 'A' is in excess.
(ii) In above reaction,
∵1 mol of A reacts with 1 mol of B
∴2 mol of A will react with 2 mol of B.
Hence, in this case, 'A' is the limiting reagent, as 'B' is in. excess.
(iii) In above reaction,
∵1 atom of A reacts with 1 molecule of B.
∴100 atoms of A will react with 100 molecules of B. Hence, in this case there is no limiting reagent.
(iv) In above reaction,
∵1 mol of 'A' reacts with 1 mol of B
∴2.5 mol of 'A' will react with 2.5 mol of B.
Hence, in this case, 'B' is limiting reagent, as 'A' is in excess.
(v) In above reaction,
∵1 mol of 'A' reacts with 1 mol of 'B'.
∴2.5 mol of 'A' will reacts with 2.5 mol of 'B'.
Hence, In this case, 'A' is limiting reagent, as, 'B' is in excess.
Question 1.24.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
Ng (g) + H_{2}(g) → 2 NH_{3} (g)
(i) Calculate the mass of ammonia produced if 2.00 10 g dinitrogen reacts with 1.00 x 10^{3} g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution:
∵ 28g of N_{2} react with 6g of Hg
∴ 1g of N reacts with g of H _{2}
∴ 2000g of Ng will react with = \(\frac{6}{28} \times 2000\)
= 428.57 g of H _{2}
Since H_{2} is in excess, so, N_{2} is the limiting reagent. N_{2} limits the amount of ammonia produced.
∵ 28g of N_{2} produces 34g of NH_{3}.
∴ 2000g of N_{2} will produce = \(\frac{34}{28}\) x 2000
= 2428.57 g of NH_{3}
(ii) Hydrogen (H_{2}) will remain unreacted because it is in excess.
(iii) Amount of H_{2} remain unreacted
= 1000 - 428.57 = 571.43 g
Question 1.25.
How are 0.50 mol Na_{2}CO_{3} and 0.50 M Na_{2}CO_{3} different?
Solution:
Molar mass of Na_{2}CO_{3 }= 2 x 23 + 12 × 3 × 16 = 10^{6} g mol^{-1}
1 mol of Na_{2}CO_{3} means 10^{6} g of Na_{2}CO_{3}
0.50 mol of Na_{2}CO_{3} means = 0.50 x 10^{6} g
= 53 g of Na_{2}CO_{3}
0.50 M Na_{2}CO_{3 }means 0.50 mol i.e., 53g of Na_{2}CO_{3} are present in 1 litre of the solution.
Question 1.26.
If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Solution:
The balanced equation is,
2H_{2}(g) + O_{2}(g) → 2H_{2}O(g)
Hence, the above equation shows that 10 volumes of water vapour[ (H_{2}O(g)] will be produced.
Question 1.27.
Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg 1pm = 10^{-12} m
Solution:
(i) ∵ 1 pm = 10^{-12} m
∴ 28.7 pm = 287 x 10^{-12} m
(ii) 15.15 pm = 15.15 x 10^{-12} m
= 15.15 x 10^{-11} m
(iii) ∵ 1mg = 10^{-6 }Kg
∴ 25365 mg = 253265 x 10^{-6} kg
= 2.5365 x 10^{-2} kg
Question 1.28.
Which one of the following will have the largest number of atoms?
(i) 1g Au(s)
(ii) 1g Na(s)
(iii) 1g Li(s)
(iv) 1g of Cl_{2}(g)
Solution:
(i) Atomic mass of Au = 197
∴ Number of atoms in 1 g Au(s) = \(\frac{1}{197}\) x 6.022 × 10^{28}
= 3.057 x 10^{21} atoms
(ii) Atomic mass of Na = 23
∴ Number of atoms in 1g Na(s) = \(\frac{1}{23}\) x 6.022 x 10^{23}
= 2.618 x 10^{22} atoms
(iii) Atomic mass of Li = 7
∴ Number of atoms in 1 g Li(s)
= \(\frac{1}{7}\) x 6022 × 10^{23}
= 8.6042 × 10^{22} atoms
(iv) Molar mass of Cl2 = 2 x 35.5 = 71 Number of atoms in 1g of Cl_{2} (g)
= \(\frac{1}{71}\) × 6022 × 10^{23}
= 8.48 x 10^{21} atoms
Therefore, 1g of Li(s) will have the largest number of atoms.
Question 1.29.
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).
Solution:
Molarity means number of moles of solute in one litre of solution.
1L of ethanol = 1 L of water
Number of moles of H_{2}O in 1 litre water
= \(\frac{1000 \mathrm{~g}}{18}\)
= 55.55 mol
For a binary solution
X1 + X2 = 1
Хн_{2}о + Хс_{2}н_{5}он = 1
∴Хн_{2}O = 1 - Хс_{2}н_{5}он
= 1 - 0.040
= 0.96
\(\begin{aligned} X_{\mathrm{H}_2 \mathrm{O}} & =\frac{n_{\mathrm{H}_2 \mathrm{O}}}{n_{\mathrm{H}_2 \mathrm{O}}+n_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}} \\ 0.96 & =\frac{55.55}{55.55+n_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}} \end{aligned}\)
or 53.328 + 0.96 nc_{2}H_{2}OH = 55.55
or
0.96 nc_{2}H_{2}OH = 55.55 - 53.328
= 2.222
\(n_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}=\frac{2.222}{0.96}\)
= 2.3145
Molarity = 2.3145 mol L^{-1}
Question 1.30.
What will be the mass of one 12C atom in g?
Solution :
1 mole of 12 C atoms = 6.022 x 10^{23}
= 12 g of carbon
∵ 6022 x 10^{23} atoms of 12 C have mass = 12 g
∴ 1 atom of 12 C will have mass
= \(\frac{12}{6.022 \times 10^{23}} g\)
= 1.9927 x 10^{-23} g atoms
Question 1.31.
How many significant figures should be present in the answer of the following calculation?
\(\text { (i) } \frac{0.02856 \times 298.15 \times 0.112}{0.5785}\)
(ii) 5 x 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Solution:
= 1648511
= 1.65
It is due to the fact that 0.112 has only three significant figures, so, final result should have three significant figures.
(ii) 5 × 5364 = 26.82
26.82 should have 4 significant figures.
(iii) 0.0125 + 0.7864 + 00215
= 0.8204
= 0.820
It is due to the fact that 0.0215 or 0.0125 has three significant figures, so, final result should have only three significant figures.
Question 1.32.
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
Istope |
Isotopic Molar mass |
Abudance |
^{36}Ar |
35.96755 g mol^{-1} |
0.377% |
^{38}Ar |
37.96272 g mol^{-1} |
0.063% |
^{40}Ar |
39.9642 g mol^{-1} |
99.600% |
Solution:
Average molar mass of argon
= (0,00337 x 3596755) + (0.00063)
= 37.96272) + (0.9960 × 39.9624)
= 0.121 + 0.024 + 39.803
= 39.948 g mol^{-1}
Question 1.33.
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
Solution:
(i) ∵ 1 mol of Ar contains 6.022 x 10^{23} atoms
∴ 52 mol of Ar will contain = 52 x 6.022 × 10^{23} atoms
= 313.144 x 10^{23} atoms
= 3.1314 x 10^{25} atoms
(ii) 4 u of He contains 1 atom of He
∴52 u of He will contain = x 52 atoms = 13 atoms
(iii) 1 mol of He = 4 g of He = 6.022 x 10^{23} atoms
= \(\frac{6.022 \times 10^{23}}{4} \times 52 \text { atoms }\)
= 7.8286 x 10^{24} atoms
Question 1.34.
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. C + O_{2} → CO_{2}
Solution:
(i) ∵44g of CO_{2} contains 12g of carbon:
∴ 3.38 g of CO_{2} will contain
= \(\frac{12}{44}\) x 3.38 g
= 0.9218 g of Carbon
∵ 18g of H_{2}O contains 2g of hydrogen
∴ 0.690 g of H_{2}O will contain
= \(\frac{2}{18}\) x 0.690 g
= 0.0767 g of hydrogen
Total mass of compound = 0.9218 + 0.0767
= 0.9985 g
% of C = \(\frac{0.9218}{0.9985}\) x 100
= 92.318
= 92.32%
% of H = \(\frac{0.0767}{0.9985}\) x 100 = 7.68%
∴ Empirical formula of compound = CH
(ii) ∵ 10.0 L of the given gas weigh 11.6 g at STP
∴ 22.4 L of the given gas will weigh
\(=\frac{11.6 \times 22.4}{10}\)
= 25984 g at STP
∴ Molar mass = 25984 g mol^{-1} = 26 g mol^{-1}
(iii) Empirical formula mass of CH = 12 + 1 = 13
= \(\frac{26}{13}\) = 2
∴ Molecular formula = n x Empirical formula
= 2 × CH
= C_{2}H_{2}
Question 1.35.
Calcium carbonate reacts with aqueous HCl to give CaCl, and CO_{2} according to the reaction,
CaCO_{3}(s) + 2HCl(aq) → CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)
What mass of CaCO_{3} is required to react completely with 25 mL of 0.75 M HCl?
Solution:
Volume of HCl = 25 mL
Molarity of HCl = 0.75 M
Molecular mass of HCl (m) = 1 + 35.5 = 36.5 g mol^{1}
= 0.6844 g
For the reaction,
CaCO_{3}(s) + 2HCl (aq) → CaCl_{2} (aq) + CO2(g) + H_{2}O(l)
100 g 2 x 36.5 = 73 g
According to balanced chemical equation,
∵ 73 g HCl reacts completely with 100 g of CaCO_{3}
∴ 0.6844 g HCl will react with
= \(\frac{100}{73}\) x 0.6844 = 0.9375 g
Question 1.36.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO_{2}) with aqueous hydrochloric acid according to the reaction
4HCl(aq) + MnO_{2} (s) → 2H_{2}O(l) + MnCl_{2} (aq) + Cl_{2}(g)
How many grams of HCl reacts with 5.0 g of manganese dioxide ?
Solution:
For the reaction,
4HCl(aq) + MnO _{2} (s) → 2H_{2}O(l) + MnCl_{2}(aq) + Cl_{2}(g)
= 146.0 g
According to above balanced chemical equation, 87g of MnO_{2} reacts with 146 g of HCl
∴5.0 g of MnO_{2} will react with
= \(\frac{146}{87} \times 5\)
= 8.39 g of HCI