NCERT Solution Class 11th Chemistry Chapter – 1 Some Basic Concepts of Chemistry Question & Answer

NCERT Solutions Class 11th Chemistry Chapter – 1 Some Basic Concepts of Chemistry

NCERT Solutions Class 11th Chemistry Chapter – 1 Some Basic Concepts of Chemistry

?Chapter – 1?

✍Excretory Products and Their Elimination✍

?Question & Answer?

Q1. Calculate the molar mass of the following:

(i) CH_{4}CH4​      (ii)H_{2}OH2​O      (iii)CO_{2}CO2​

Answer:

(i) CH_{4}CH4​ :

Molecular mass of CH_{4}CH4​ = Atomic mass of C + 4 x Atomic mass of H

= 12 + 4 x 1

= 16 u

(ii) H_{2}OH2​O :

Molar mass of water H_{2}OH2​O

Atomic mass of H = 1

Atomic mass of O = 16

H₂O = 2×H+1×O

Molar mass of water = 2×1+16 = 18g/mol

(iii) CO_{2}CO2​ :

Molecular mass of CO_{2}CO2​ = Atomic mass of C + 2 x Atomic mass of O

= 12 + 2 × 16

= 44 u

Q2. Calculate the mass per cent of different elements present in sodium sulphate (Na_{2}SO_{4}Na2​SO4​) .

Answer:

Now for Na_{2}SO_{4}Na2​SO4​.

Molar mass of Na_{2}SO_{4}Na2​SO4​

= [(2 x 23.0) + (32.066) + 4(16.00)]

=142.066 g

Formula to calculate mass percent of an element = \frac{Mass\;of\;that\;element\;in\;the\;compound}{Molar\;mass\;of\;the\;compound}\times 100MolarmassofthecompoundMassofthatelementinthecompound​×100

Therefore, mass percent of the sodium element:

= \frac{46.0g}{142.066g}\times 100142.066g46.0g​×100

= 32.379

= 32.4%

Mass percent of the sulphur element:

= \frac{32.066g}{142.066g}\times 100142.066g32.066g​×100

= 22.57

= 22.6%

Mass percent of the oxygen element:

= \frac{64.0g}{142.066g}\times 100142.066g64.0g​×100

= 45.049

= 45.05%

Q3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Answer:  Given there is an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass:

Relative moles of iron in iron oxide:

= \frac{percent\;of\;iron\;by\;mass}{Atomic\;mass\;of\;iron}Atomicmassofironpercentofironbymass​

= \frac{69.9}{55.85}55.8569.9​

= 1.25

Relative moles of oxygen in iron oxide:

= \frac{percent\;of\;oxygen\;by\;mass}{Atomic\;mass\;of\;oxygen}Atomicmassofoxygenpercentofoxygenbymass​

= \frac{30.1}{16.00}16.0030.1​

= 1.88

The simplest molar ratio of iron to oxygen:

⇒ 1.25: 1.88 ⇒ 1: 1.5 ⇒ 2: 3

Therefore, the empirical formula of the iron oxide is Fe_{2}O_{3}Fe2​O3​.

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.

Answer:

(i) 1 mole of carbon is burnt in air.

C+O_{2}\rightarrow CO_{2}C+O2​→CO2​

1 mole of carbon reacts with 1 mole of O₂ to form one mole of CO₂.

Amount of CO_{2}CO2​ produced = 44 g

(ii) 1 mole of carbon is burnt in 16 g of O₂.

1 mole of carbon burnt in 32 grams of O₂ it forms 44 grams of CO_{2}CO2​.

Therefore, 16 grams of O₂ will form \frac{44\times 16}{32}3244×16​

= 22 grams of CO_{2}CO2​

(iii) 2 moles of carbon are burnt in 16 g of O₂.

Here again, dioxygen is the limiting reactant. 16g of dioxygen can combine only with 0.5mol of carbon. CO₂ produced again is equal to 22g.

Q5. Calculate the mass of sodium acetate (CH_{3}COONa)(CH3​COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^–1.

Answer:

0.375 M aqueous solution of CH_{3}COONaCH3​COONa

= 1000 mL of solution containing 0.375 moles of CH_{3}COONaCH3​COONa

Therefore, no. of moles of CH_{3}COONaCH3​COONa in 500 mL

= \frac{0.375}{1000}\times 50010000.375​×500

= 0.1875 mole

Molar mass of sodium acetate = 82.0245\;g\;mol^{-1}82.0245gmol−1

Therefore, the mass of CH_{3}COONaCH3​COONa

= (82.0245\;g\;mol^{-1})(0.1875\;mole)(82.0245gmol−1)(0.1875mole)

= 15.38 grams

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%

Answer:

Mass percent of HNO₃ in sample is 69 %

Thus, 100 g of HNO₃ contains 69 g of HNO₃ by mass.

Molar mass of HNO₃

= { 1 + 14 + 3(16)} g.mol^{-1}g.mol−1

= 1 + 14 + 48

= 63g\;mol^{-1}=63gmol−1

Now, no. of moles in 69 g of HNO_{3}HNO3​:

= \frac{69\:g}{63\:g\:mol^{-1}}63gmol−169g​

= 1.095 mol

Volume of 100g HNO₃ solution

= \frac{Mass\;of\;solution}{density\;of\;solution}densityofsolutionMassofsolution​

= \frac{100g}{1.41g\;mL^{-1}}1.41gmL−1100g​

= 70.2mL

= 70.92\times 10^{-3}\;L70.92×10−3L

Concentration of HNO₃

= \frac{1.095\:mole}{70.92\times 10^{-3}L}70.92×10−3L1.095mole​

= 15.44mol/L

Therefore,

Concentration of HNO₃ = 15.44 mol/L

Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Answer:

1 mole of CuSO_{4}CuSO4​ contains 1 mole of Cu.

Molar mass of CuSO_{4}CuSO4​

= (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 grams

159.5 grams of CuSO_{4}CuSO4​ contains 63.5 grams of Cu.

Therefore, 100 grams of CuSO_{4}CuSO4​ will contain \frac{63.5\times 100g}{159.5}159.563.5×100g​ of Cu.

= \frac{63.5\times 100}{159.5}159.563.5×100​

=39.81 grams

Q8. 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. 

Answer: Here,

Mass percent of Fe = 69.9%

Mass percent of O = 30.1%

No. of moles of Fe present in oxide

= \frac{69.90}{55.85}55.8569.90​

= 1.25

No. of moles of O present in oxide

= \frac{30.1}{16.0}16.030.1​

=1.88

Ratio of Fe to O in oxide,

= 1.25: 1.88

= \frac{1.25}{1.25}:\frac{1.88}{1.25}1.251.25​:1.251.88​

=$1:1.5$

= $2:3$

Therefore, the empirical formula of oxide is Fe_{2}O_{3}Fe2​O3​

Empirical formula mass of Fe_{2}O_{3}Fe2​O3​

= [2(55.85) + 3(16.00)] g

= 159. 7g

The molar mass of Fe_{2}O_{3}Fe2​O3​ = 159.69g

Therefore n = \frac{Molar\;mass}{Empirical\;formula\;mass}=\frac{159.69\;g}{159.7\;g}EmpiricalformulamassMolarmass​=159.7g159.69g​

= 0.999

= 1(approx)

The molecular formula of a compound can be obtained by multiplying n with the empirical formula.

Thus, the empirical of the given oxide is Fe_{2}O_{3}Fe2​O3​ and n is 1.

Therefore, the molecular formula of the oxide is Fe_{2}O_{3}Fe2​O3​

Q9. Calculate the atomic mass (average) of chlorine using the following data:

Answer: Fractional Abundance of ³⁵Cl = 0.7577 and Molar mass = 34.9689

Fractional Abundance of ³⁷Cl = 0.2423 and Molar mass = 36.9659

Average Atomic mass = (0.7577 x 34.9689)amu + (0.2423 x 36.9659)

= 26.4959 + 8.9568 = 35.4527

Q10. In three moles of ethane (C₂H₆), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atom
(iii) Number of molecules of ethane

Answer:

(i) 1 mole of C_{2}H_{6}C2​H6​ contains two moles of C- atoms.

$\therefore$ No. of moles of C- atoms in 3 moles of C_{2}H_{6}C2​H6​.

= 2 x 3

= 6

(ii) 1 mole of C_{2}H_{6}C2​H6​ contains six moles of H- atoms.

$\therefore$ No. of moles of H- atoms in 3 moles of C_{2}H_{6}C2​H6​.

= 3 x 6

= 18

(iii) 1 mole of C_{2}H_{6}C2​H6​ contains 1 mole of ethane- atoms.

$\therefore$ No. of molecules in 3 moles of C_{2}H_{6}C2​H6​.

= 3 x 6.023 x 10^{23}1023

= 18.069 x 10^{23}1023

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 2L?

Answer: Molarity (M) is as given by,

= \frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}VolumeofsolutioninLitresNumberofmolesofsolute​

= \frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}2LMolarmassofsugarMassofsugar​​

= \frac{\frac{20\;g}{[(12\;\times \;12)\;+\;(1\;\times \;22)\;+\;(11\;\times \;16)]g]}}{2\;L}2L[(12×12)+(1×22)+(11×16)]g]20g​​

= \frac{\frac{20\;g}{342\;g}}{2\;L}2L342g20g​​

= \frac{0.0585\;mol}{2\;L}2L0.0585mol​

= 0.02925 molL^{-1}L−1

Therefore, Molar concentration = 0.02925 molL^{-1}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?

Answer: Molar mass of methanol (CH₃OH)

= 32 gmol^-1 = 0.032 kgmol^-1

molarity of the given solution

=\frac{W_{2}in kg}{M_{w_{2}}\times V_{(sol)}L}=\frac{d_{sol}(kgL^{-1})}{Mw_{2}(kg)}\\=\frac{0.793kgL^{-1}}{0.032kgmol^{-1}}= 24.78 M\\\underset{(Given solution)}{Applying M_{1}\times V_{1}}= \underset{(solution to be prepared)}{M_{2}V_{2}}=Mw2​​×V(sol)​LW2​inkg​=Mw2​(kg)dsol​(kgL−1)​=0.032kgmol−10.793kgL−1​=24.78M(Givensolution)ApplyingM1​×V1​​=(solutiontobeprepared)M2​V2​​

24.78 x V₁ = 0.25 x 2.5 L

or V₁ = 0.02522L = 25.22mL

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

Answer: Pressure is the force (i.e., weight) acting  per unit area  But weight = mg

∴ Pressure = Weight per unit area

=\frac{1034g\times 9.8ms^{-2}}{cm^{2}}\\=\frac{1034g\times 9.8ms^{-2}}{cm^{2}}\times \frac{1kg}{1000g}\times \frac{100cm\times 100cm}{1m\times 1m}\times \frac{1N}{kgms^{-2}}\times \frac{1Pa}{1Nm^{-2}}\\= 1.01332\times 10^{^{5}}Pa=cm21034g×9.8ms−2​=cm21034g×9.8ms−2​×1000g1kg​×1m×1m100cm×100cm​×kgms−21N​×1Nm−21Pa​=1.01332×105Pa

Q14. What is the SI unit of mass? How is it defined?

Answer: The S.I unit of mass is kilogram (kg). A kilogram is equal to the mass of a platinum-iridium cylinder kept at the International Bureau of Weights and Measures at Service, France.

Q15. Match the following prefixes with their multiples:

Answer:

Q16. What do you mean by significant figures?

Answer:

Significant figures are the meaningful digits which are known with certainty. Significant figures indicate uncertainty in experimented value.

e.g.: The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. The total significant figures are 3.

Therefore, “the total number of digits in a number with the last digit that shows the uncertainty of the result is known as significant figures.”

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 per cent by mass.
(ii) Determine the molality of chloroform in the water sample.

Answer:

(i) 1 ppm = 1 part out of 1 million parts.

Mass percent of 15 ppm chloroform in H₂O

= \frac{15}{10^{6}} \times 10010615​×100

= $\approx$ 1.5 ×10^{-3}10−3 %

(ii) Molarity = \frac{15/119.5}{10^{6}\times 10^{-3}}= 1.25 \times 10^{-4}Molarity=106×10−315/119.5​=1.25×10−4

Q18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Answer:

(i) 0.0048= 4.8 ×10^{-3}10−3

(ii) 234,000 = 2.34 ×10^{5}105

(iii) 8008= 8.008 ×10^{3}103

(iv) 500.0 = 5.000 ×10^{2}102

(v) 6.0012 = 6.0012 ×10^{0}100

Q19. How many significant figures are present in the following?
(a) 0.0025
(b) 208
(c) 5005
(d) 126,000
(e) 500.0
(f) 2.0034

Answer:

(a) 0.0025: 2 significant numbers.

(b) 208: 3 significant numbers.

(c) 5005: 4 significant numbers.

(d) 126,000:3 significant numbers.

(e) 500.0: 4 significant numbers.

(f) 2.0034: 5 significant numbers.

Q20. Round up the following upto three significant figures:
(a) 34.216
(b) 10.4107
(c)0.04597
(d)2808

Answer:

(a) The number after round up is: 34.2

(b) The number after round up is: 10.4

(c)The number after round up is: 0.0460

(d)The number after round up is: 2810

Q21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

(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

Answer:

(a)

Here if we fix the mass of dinitrogen at 14g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 16g, 32g, 32g, and 80g.

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.

(b)

 (i.) 1 km = 1 km \times \frac{1000m}{1km}\times \frac{100cm}{1m}\times \frac{10mm}{1cm}= 10^{6}mm\\1km = 1 km \times \frac{1000m}{1km}\times \frac{1pm}{10^{-12}m}= 10^{^{15}}pm1km=1km×1km1000m​×1m100cm​×1cm10mm​=106mm1km=1km×1km1000m​×10−12m1pm​=1015pm

(ii.) 1 mg = 1 mg\times \frac{1g}{1000mg}\times \frac{1kg}{1000g}= 10^{-6}kg\\1mg = 1mg\times \frac{1g}{1000mg}\times \frac{1ng}{10^{-9}g}=10^{6}ng1mg=1mg×1000mg1g​×1000g1kg​=10−6kg1mg=1mg×1000mg1g​×10−9g1ng​=106ng

(iii.) 1mL = 1mL\times \frac{1L}{1000mL}=10^{-3}L\\1mL = 1cm^{3} \\ =1cm^{3}\times \frac{1dm\times 1dm\times 1dm}{10cm\times 10cm\times 10cm}= 10^{-3}dm^{3}1mL=1mL×1000mL1L​=10−3L1mL=1cm3=1cm3×10cm×10cm×10cm1dm×1dm×1dm​=10−3dm3

Q22. If the speed of light is 3.0 × 10⁸ m s^–1, calculate the distance covered by light in 2.00 ns

Answer: Time taken = 2 ns

= 2 × 10^{ -9 }10−9 s

Now,

Speed of light = 3 × 10^{ 8 }108 ms^{ -1 }ms−1

We know that,

Distance = Speed x Time

So,

Distance travelled in 2 ns = speed of light x time taken

= (3 × 10^{ 8 }108)(2 × 10^{ -9 }10−9)

= 6 × 10^{ -1 }10−1 m

= 0.6 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

Answer: Limiting reagent: It determines the extent of a reaction. It is the first to get consumed during a reaction, thus causes the reaction to stop and limits the amount of product formed.

(i) 300 atoms of A + 200 molecules of B

1 atom of A reacts with 1 molecule of B. Similarly, 200 atoms of A reacts with 200 molecules of B, so 100 atoms of A are unused. Hence, B is the limiting reagent.

(ii) 2 mol A + 3 mol B

1 mole of A reacts with 1 mole of B. Similarly, 2 moles of A reacts with 2 moles of B, so 1 mole of B is unused. Hence, A is the limiting reagent.

(iii) 100 atoms of A + 100 molecules of Y

1 atom of A reacts with 1 molecule of Y. Similarly, 100 atoms of A reacts with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting reagent.

(iv) 5 mol A + 2.5 mol B

1 mole of A reacts with 1 mole of B. Similarly 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of A is unused. Hence, B is the limiting reagent.

(v) 2.5 mol A + 5 mol B

1 mole of A reacts with 1 mole of B. Similarly, 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of B is unused. 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 NH_{ 3 }NH3​ produced if 2 \; \times \;10^{ 3 }2×103 g N₂ reacts with 1 \; \times \;10^{ 3 }1×103 g of H₂?

 (ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass.

Answer:

(i) 1 mol of N₂ i.e., 28 g reacts with 3 moles of H₂ i.e., 6 g of H₂

∴ 2000 g of N₂ will react with H₂ = \frac{6}{28}\times 200g = 428.6g286​×200g=428.6g

Thus, N₂ is the limiting reagent while H₂ is the excess reagent

2 mol of N₂ i.e., 28 g of N₂ produces NH₃ = 2 mol

= 34 g

Therefore, 2000 g will produces NH₃ = \frac{34}{28}\times 2000 g2834​×2000g

= 2428.57 g

(ii) H₂ will remain unreacted

(iii) Mass left unreacted = 1000g – 428.6g = 571.4g

Q25. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Answer:

Molar mass of Na_{ 2 }CO_{ 3 }Na2​CO3​:

= (2 × 23) + 12 + (3 × 16)

= 106 g mol^{ -1 }mol−1

1 mole of Na_{ 2 }CO_{ 3 }Na2​CO3​ means 106 g of Na_{ 2 }CO_{ 3 }Na2​CO3​

Therefore, 0.5 mol of Na_{ 2 }CO_{ 3 }Na2​CO3​

= \frac{ 106 \; g }{ 1 \; mol } \; \times \; 0.5 \; mol1mol106g​×0.5mol Na_{ 2 }CO_{ 3 }Na2​CO3​

= 53 g of Na_{ 2 }CO_{ 3 }Na2​CO3​

0.5 M of Na_{ 2 }CO_{ 3 }Na2​CO3​ = 0.5 mol/L Na_{ 2 }CO_{ 3 }Na2​CO3​

Hence, 0.5 mol of Na_{ 2 }CO_{ 3 }Na2​CO3​ is in 1 L of water or 53 g of Na_{ 2 }CO_{ 3 }Na2​CO3​ is in 1 L of water.

Q26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer: Reaction:

2H_{ 2 }\;(g)  \; + \; O_{ 2 }\; (g)  \; \rightarrow \; 2H_{ 2 }O\; (g)2H2​(g) +O2​(g) →2H2​O(g)

2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of water vapour.

Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of water vapour.

Q27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Answer:

(i) 28.7 pm

1 pm = 10^{ -12 } \; m10−12m

28.7 pm = 28.7 × 10^{ -12 } \; m10−12m

= 2.87 × 10^{ -11 } \; m10−11m

(ii) 15.15 pm

1 pm = 10^{ -12 } \; m10−12m

15.15 pm = 15.15 × 10^{ -12 } \; m10−12m

= 1.515 × 10^{ -11 } \; m10−11m

(iii) 25365 mg

1 mg = 10^{ -3 } \; g10−3g

1 mg = 10^-6 kg

25365 mg = 25365 x 10^-6 kg

25365 mg = 2.5365 × 10^{ -2 } \; kg10−2kg

Q28. Which one of the following will have the largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl_{ 2 }Cl2​ (g)

Answer:

(i) 1 g of Au (s)

= \frac{ 1 }{ 197 }1971​ mol of Au (s)

= \frac{ 6.022 \; \times \; 10^{ 23 } }{ 197 }1976.022×1023​ atoms of Au (s)

= 3.06 \times \; 10^{ 21 }×1021 atoms of Au (s)

(ii) 1 g of Na (s)

= \frac{ 1 }{ 23 }231​ mol of Na (s)

= \frac{ 6.022 \; \times \; 10^{ 23 } }{ 23 }236.022×1023​ atoms of Na (s)

= 0.262 \times \; 10^{ 23 }×1023 atoms of Na (s)

= 26.2 \times \; 10^{ 21 }×1021 atoms of Na (s)

(iii) 1 g of Li (s)

= \frac{ 1 }{ 7 }71​ mol of Li (s)

= \frac{ 6.022 \; \times \; 10^{ 23 } }{ 7 }76.022×1023​ atoms of Li (s)

= 0.86 \times \; 10^{ 23 }×1023 atoms of Li (s)

= 86.0 \times \; 10^{ 21 }×1021 atoms of Li (s)

(iv)1 g of Cl_{ 2 }Cl2​ (g)

= \frac{ 1 }{ 71 }711​ mol of Cl_{ 2 }Cl2​ (g)

(Molar mass of Cl_{ 2 }Cl2​ molecule = 35.5 × 2 = 71 g mol^{ -1 }mol−1)

= \frac{ 6.022 \; \times \; 10^{ 23 } }{ 71 }716.022×1023​ atoms of Cl_{ 2 }Cl2​ (g)

= 0.0848 \times \; 10^{ 23 }×1023 atoms of Cl_{ 2 }Cl2​ (g)

= 8.48 \times \; 10^{ 21 }×1021 atoms of Cl_{ 2 }Cl2​ (g)

Therefore, 1 g of Li (s) will have the largest no. 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).

Answer:  Mole fraction of C_{ 2 }H_{ 5 }OHC2​H5​OH

= \frac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}NumberofmolesofsolutionNumberofmolesofC2​H5​OH​

0.040 = \frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; n_{H_{ 2 }O}}nC2​H5​OH​+nH2​O​nC2​H5​OH​​ ——(1)

No. of moles present in 1 L water:

n{ H_{ 2 }O} \; = \; \frac{ 1000 \; g}{18 \; g \; mol^{ -1 }}nH2​O​=18gmol−11000g​ n_{ H_{ 2 }O}nH2​O​ = 55.55 mol

Substituting the value of n_{ H_{ 2 }O}nH2​O​ in eq (1),

\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; 55.55}nC2​H5​OH​+55.55nC2​H5​OH​​ = 0.040

n_{C_{ 2 }H_{ 5 }OH}nC2​H5​OH​ = 0.040n_{C_{ 2 }H_{ 5 }OH}nC2​H5​OH​ + (0.040)(55.55)

0.96n_{C_{ 2 }H_{ 5 }OH}nC2​H5​OH​ = 2.222 mol

n_{C_{ 2 }H_{ 5 }OH}nC2​H5​OH​ = \frac{ 2.222 }{ 0.96 } \; mol0.962.222​mol n_{C_{ 2 }H_{ 5 }OH}nC2​H5​OH​ = 2.314 mol

Therefore, molarity of solution

= \frac{ 2.314 \; mol }{ 1 \; L }1L2.314mol​

= 2.314 M

Q30. What will be the mass of one ¹²C atom in g?

Answer: 1 mole of carbon atoms

= 6.023 \; \times \; 10^{ 23 }6.023×1023 atoms of carbon

= 12 g of carbon

Therefore, mass of 1 atom of _{}^{ 12 }\textrm{ C }12​ C 

= \frac{ 12 \; g }{ 6.022 \; \times \; 10^{ 23 }}6.022×102312g​

= 1.993 \; \times \; 10^{ -23 } g1.993×10−23g

Q31. How many significant figures should be present in the answer of the following calculations?

(i) \frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }0.57850.02856×298.15×0.112​

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

Answer:

(i) \frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }0.57850.02856×298.15×0.112​

Least precise number = 0.112

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 0.112

= 3

(ii) 5 × 5.364

Least precise number = 5.364

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 5.364

= 4

(iii) 0.0125 + 0.7864 + 0.0215 As the least no. of decimal place in each term is 4. Hence, the no. of significant numbers in the answer is also 4.

Q32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Answer: Molar mass of Argon:

= [( 35.96755 \; \times \; \frac{ 0.337 }{ 100 })(35.96755×1000.337​) + ( 37.96272 \; \times \; \frac{ 0.063 }{ 100 })(37.96272×1000.063​) + ( 39.9624 \; \times \; \frac{ 99.600 }{ 100 })(39.9624×10099.600​)]

= [0.121 + 0.024 + 39.802] g \; mol^{ -1 }gmol−1

= 39.947 g \; mol^{ -1 }gmol−1

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

Answer:

(i) 52 moles of Ar

1 mole of Ar = 6.023 \; \times \; 10^{ 23 }6.023×1023 atoms of Ar

Therefore, 52 moles of Ar = 52 × 6.023 \; \times \; 10^{ 23 }6.023×1023 atoms of Ar

= 3.131 \; \times \; 10^{ 25 }3.131×1025 atoms of Ar

(ii) 52 u of He

1 atom of He = 4 u of He

OR

4 u of He = 1 atom of He

1 u of He = \frac{ 1 }{ 4 }41​ atom of He

52 u of He = \frac{ 52 }{ 4 }452​ atom of He

= 13 atoms of He

(iii) 52 g of He

4 g of He = 6.023 \; \times \; 10^{ 23 }6.023×1023 atoms of He

52 g of He = \frac{ 6.023 \; \times \; 10^{ 23 } \; \times \;52 }{ 4 }46.023×1023×52​ atoms of He

= 7.829 \; \times \; 10^{ 24 }7.829×1024 atoms of He

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.  Find:
(i) Empirical formula
(ii) Molar mass of the gas, and
(iii) Molecular formula

Answer:

(i) Empirical formula

1 mole of CO_{ 2 }CO2​ contains 12 g of carbon

Therefore, 3.38 g of CO_{ 2 }CO2​ will contain carbon

= \frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g44g12g​×3.38g

= 0.9218 g

18 g of water contains 2 g of hydrogen

Therefore, 0.690 g of water will contain hydrogen

= \frac{ 2 \; g }{ 18 \; g } \; \times 0.69018g2g​×0.690

= 0.0767 g

As hydrogen and carbon are the only elements of the compound. Now, the total mass is:

= 0.9217 g + 0.0767 g

= 0.9984 g

Therefore, % of C in the compound

= \frac{ 0.9217 \; g }{ 0.9984 \; g } \; \times 1000.9984g0.9217g​×100

= 92.32 %

% of H in the compound

= \frac{ 0.0767 \; g }{ 0.9984 \; g } \; \times 1000.9984g0.0767g​×100

= 7.68 %

Moles of C in the compound,

= \frac{ 92.32 }{ 12.00 }12.0092.32​

= 7.69

Moles of H in the compound,

= \frac{ 7.68 }{ 1 }17.68​

= 7.68

Therefore, the ratio of carbon to hydrogen is,

7.69: 7.68

1: 1

Therefore, the empirical formula is CH.

(ii) Molar mass of the gas

Weight of 10 L of gas at STP = 11.6 g

Therefore, weight of 22.4 L of gas at STP

= \frac{ 11.6 \; g }{ 10 \; L } \; \times \; 22.4 \; L10L11.6g​×22.4L

= 25.984 g

$\approx$ 26 g

(iii) Molecular formula

Empirical formula mass:

CH = 12 + 1

= 13 g

n = \frac{ Molar \; mass \; of \; gas}{Empirical \; formula \; mass \; of \; gas}EmpiricalformulamassofgasMolarmassofgas​

= \frac{ 26 \; g }{ 13 \; g}13g26g​

= 2

Therefore, molecular formula = 2 x CH = C_{ 2 }H_{ 2 }C2​H2​.

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?

Answer:

0.75 M of HCl

≡ 0.75 mol of HCl are present in 1 L of water

≡ [(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 27.375 g of HCl

Therefore, amt of HCl present in 25 mL of solution

= \frac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL1000mL27.375g​×25mL

= 0.6844 g

Given chemical reaction,

CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq)  \; \rightarrow \; CaCl_{ 2 }\;(aq)  \; + \; CO_{ 2 }\; (g)  \; + \; H_{ 2 }O\; (l)CaCO3​(s)+2HCl(aq) →CaCl2​(aq) +CO2​(g) +H2​O(l)

2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO_{ 3 }CaCO3​ (100 g)

Therefore, amt of CaCO_{ 3 }CaCO3​ that will react with 0.6844 g

= \frac{ 100 }{ 73 } \; \times \; 0.6844 \; g73100​×0.6844g

= 0.9375 g

Q36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction: 4 HCl (aq) + MnO₂(s) → 2H₂O (l) + MnCl₂(aq) + Cl₂ (g) How many grams of HCl react with 5.0 g of manganese dioxide?

Answer:

1 mole of MnO_{2}MnO2​ = 55 + 2 × 16 = 87 g

4 mole of HCl = 4 × 36.5 = 146 g

1 mole of MnO_{2}MnO2​ reacts with 4 mol of HCl

Hence,

5 g of MnO_{ 2 }MnO2​will react with:

= \frac{146 \; g}{87 \; g} \; \times \; 5 \; g87g146g​×5g HCl

= 8.4 g HCl

Therefore, 8.4 g of HCl will react with 5 g of MnO_{2}MnO2​.