# NCERT Solutions For Class 11 Chemistry Chapter 1

## NCERT Solutions Class 11 Chemistry Some Basic Concepts of Chemistry

NCERT solutions class 11 chemistry chapter 1 some basic concepts chemistry is one of the key topics asked in class 11 chemistry examination. The NCERT solutions for class 11 chemistry chapter 1 some basic concepts chemistry is given here so that 11th standard candidates can prepare for their examination in an easy and interactive way. NCERT solutions for class 11 chemistry chapter 1 is prepared by subject experts according to the latest syllabus of the Central Board of Secondary Education (CBSE). The NCERT solutions for class 11 chemistry chapter 1 pdf is provided below.

Exercise

Q1. Find out the value of molecular weight of the given compounds:

(i) CH4$CH_{4}$      (ii)H2O$H_{2}O$      (iii)CO2$CO_{2}$

Ans.

(i)CH4$CH_{4}$ :

Molecular weight of methane, CH4$CH_{4}$

= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)

= [1(12.011 u) +4 (1.008u)]

= 12.011u + 4.032 u

= 16.043 u

(ii) H2O$H_{2}O$ :

Molecular weight of water, H2O$H_{2}O$

= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)

= [2(1.0084) + 1(16.00 u)]

= 2.016 u +16.00 u

= 18.016u

So approximately

= 18.02 u

(iii) CO2$CO_{2}$ :

= Molecular weight of carbon dioxide, CO2$CO_{2}$

= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)

= [1(12.011 u) + 2(16.00 u)]

= 12.011 u +32.00 u

= 44.011 u

So approximately

= 44.01u

Q2. Sodium Sulphate (Na2SO4$Na_{2}SO_{4}$) has various elements, find out the mass percentage of each element.

Ans.

Now for Na2SO4$Na_{2}SO_{4}$.

Molar mass of Na2SO4$Na_{2}SO_{4}$

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

=142.066 g

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

Therefore, Mass percent of the sodium element:

= 46.0g142.066g×100$\frac{46.0g}{142.066g}\times 100$

= 32.379

=32.4%

Mass percent of the sulphur element:

=32.066g142.066g×100$\frac{32.066g}{142.066g}\times 100$

= 22.57

=22.6%

Mass percent of the oxygen element:

= 64.0g142.066g×100$\frac{64.0g}{142.066g}\times 100$

=45.049

=45.05%

Q3. Find out the empirical formula of an oxide of iron having 69.9% Fe and 30.1% O2 by mass.

Ans.

Percent of Fe by mass = 69.9 % [As given above]

Percent of O2 by mass = 30.1 % [As given above]

Relative moles of Fe in iron oxide:

= percentofironbymassAtomicmassofiron$\frac{percent\;of\;iron\;by\;mass}{Atomic\;mass\;of\;iron}$

= 69.955.85$\frac{69.9}{55.85}$

= 1.25

Relative moles of O in iron oxide:

= percentofoxygenbymassAtomicmassofoxygen$\frac{percent\;of\;oxygen\;by\;mass}{Atomic\;mass\;of\;oxygen}$

= 30.116.00$\frac{30.1}{16.00}$

= 1.88

Simplest molar ratio of Fe to O:

= 1.25: 1.88

= 1: 1.5

$\approx$ 2: 3

Therefore, empirical formula of iron oxide is Fe2O3$Fe_{2}O_{3}$.

Q4. Find out the amount of CO2 that can be produced when

(i) 1 mole carbon is burnt in air.

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

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

Ans.

(i) 1 mole of carbon is burnt in air.

C+O2CO2$C+O_{2}\rightarrow CO_{2}$

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

Amount of CO2$CO_{2}$ 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 CO2$CO_{2}$.

Therefore, 16 grams of O2 will form 44×1632$\frac{44\times 16}{32}$

= 22 grams of CO2$CO_{2}$

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

If 1 mole of carbon are burnt in 16grams of O2 it forms 22 grams of CO2$CO_{2}$

Therefore, if 2 moles of carbon are burnt it will form

= 2×221$\frac{2\times 22}{1}$

= 44g of CO2$CO_{2}$

Q5. Find out the mass of CH3COONa$CH_{3}COONa$(sodium acetate) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of CH3COONa$CH_{3}COONa$ is 82.0245gmol1$82.0245\;g\;mol^{-1}$

Ans.

0.375 Maqueous solution of CH3COONa$CH_{3}COONa$

= 1000 mL of solution containing 0.375 moles of CH3COONa$CH_{3}COONa$

Therefore, no. of moles of CH3COONa$CH_{3}COONa$ in 500 mL

= 0.3751000×1000$\frac{0.375}{1000}\times 1000$

= 0.1875 mole

Molar mass of sodium acetate = 82.0245gmol1$82.0245\;g\;mol^{-1}$

Therefore, mass that is required of CH3COONa$CH_{3}COONa$

= (82.0245gmol1)(0.1875mole)$(82.0245\;g\;mol^{-1})(0.1875\;mole)$

= 15.38 gram

Q6. A sample of HNO3 has a density of 1.41gmL1$1.41\;g\;mL^{-1}$ find the concentration of HNO3 in moles per litre and the mass percent of HNO3 in it is 69%.

Ans.

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)} gmol1$g\;mol^{-1}$

= 1 + 14 + 48

=63gmol1$= 63g\;mol^{-1}$

Now, No. of moles in 69 g of HNO3$HNO_{3}$:

= 69g63gmol1$\frac{69\:g}{63\:g\:mol^{-1}}$

= 1.095 mol

Volume of 100g HNO3 solution

= Massofsolutiondensityofsolution$\frac{Mass\;of\;solution}{density\;of\;solution}$

= 100g1.41gmL1$\frac{100g}{1.41g\;mL^{-1}}$

= 70.92mL

= 70.92×103L$70.92\times 10^{-3}\;L$

Concentration of HNO3

= 1.095mole70.92×103L$\frac{1.095\:mole}{70.92\times 10^{-3}L}$

= 15.44mol/L

Therefore, Concentration of HNO3 = 15.44 mol/L

Q7. How much Cu (Copper) can be obtained from 100 gram of CuSO4$CuSO_{4}$( copper sulphate)?

Ans.

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

Molar mass of CuSO4$CuSO_{4}$

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

= 63.5 + 32.00 + 64.00

= 159.5 gram

159.5 gram of CuSO4$CuSO_{4}$ contains 63.5 gram of Cu.

Therefore, 100 gram of CuSO4$CuSO_{4}$ will contain 63.5×100g159.5$\frac{63.5\times 100g}{159.5}$ of Cu.

= 63.5×100159.5$\frac{63.5\times 100}{159.5}$

=39.81 gram

Q8. The mass percent of iron and oxygen in an oxide of iron is 69.9 and 30.1 calculate the molecular formula of the oxide of iron. 159.69gmol1$159.69\;g\;mol^{-1}$ is the given molar mass of an oxide.

Ans.

Here,

Mass percent of Fe = 69.9%

Mass percent of O = 30.1%

No. of moles of Fe present in oxide

= 69.9055.85$\frac{69.90}{55.85}$

= 1.25

No. of moles of O present in oxide

= 30.116.0$\frac{30.1}{16.0}$

=1.88

Ratio of Fe to O in oxide,

= 1.25: 1.88

= 1.251.25:1.881.25$\frac{1.25}{1.25}:\frac{1.88}{1.25}$

=1:1.5$1:1.5$

= 2:3$2:3$

Therefore, the empirical formula of oxide is Fe2O3$Fe_{2}O_{3}$

Empirical formula mass of Fe2O3$Fe_{2}O_{3}$

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

= 159.69 g

Therefore n = MolarmassEmpiricalformulamass=159.69g159.7g$\frac{Molar\;mass}{Empirical\;formula\;mass}=\frac{159.69\;g}{159.7\;g}$

= 0.999

= 1(approx)

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

Thus, the empirical of the given oxide is Fe2O3$Fe_{2}O_{3}$ and n is 1.

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

 Percentage Natural Abundance Molar Mass 35Cl$_{}^{35}\textrm{Cl}$ 75.77 34.9689 37Cl$_{}^{37}\textrm{Cl}$ 24.23 36.9659

Ans.

Average atomic mass of Cl.

=[(Fractional abundance of 35Cl$_{}^{35}\textrm{Cl}$)(molar mass of 35Cl$_{}^{35}\textrm{Cl}$)+(fractional abundance of 37Cl$_{}^{37}\textrm{Cl}$ )(Molar mass of 37Cl$_{}^{37}\textrm{Cl}$ )]

=[{(75.77100(34.9689u)$\frac{75.77}{100}(34.9689u)$ } + {(24.23100(34.9659u)$\frac{24.23}{100}(34.9659\;u)$ }]

= 26.4959 + 8.9568

= 35.4527 u

Therefore, the average atomic mass of Cl = 35.4527 u

Q10. In 3 moles of ethane (C2H6$C_{2}H_{6}$), calculate the given below:

(a) No. of moles of C- atoms

(b) No. of moles of H- atoms.

(c) No. of molecules of C2H6.

Ans.

(a) 1 mole C2H6$C_{2}H_{6}$ contains two moles of C- atoms.

$∴$ No. of moles of C- atoms in 3 moles of C2H6$C_{2}H_{6}$.

= 2 * 3

= 6

(b) 1 mole C2H6$C_{2}H_{6}$ contains six moles of H- atoms.

$∴$ No. of moles of C- atoms in 3 moles of C2H6$C_{2}H_{6}$.

= 3 * 6

= 18

(c) 1 mole C2H6$C_{2}H_{6}$ contains six moles of H- atoms.

$∴$ No. of molecules in 3 moles of C2H6$C_{2}H_{6}$.

= 3 * 6.023 * 1023$10^{23}$

= 18.069 * 1023$10^{23}$

Q11. What is the concentration of sugar C12H22O11$C_{12}H_{22}O_{11}$ in mol L1$L^{-1}$ if its 20 gram are dissolved in enough H2O to make a final volume up to 2 Litre?

Ans.

Molarity (M) is as given by,

= NumberofmolesofsoluteVolumeofsolutioninLitres$\frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}$

= MassofsugarMolarmassofsugar2L$\frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}$

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

= 20g342g2L$\frac{\frac{20\;g}{342\;g}}{2\;L}$

= 0.0585mol2L$\frac{0.0585\;mol}{2\;L}$

= 0.02925 molL1$L^{-1}$

Therefore, Molar concentration = 0.02925 molL1$L^{-1}$

Q12. The density of CH3OH (methanol) is 0.793 kg L1$L^{-1}$. For making 2.5 Litre of its 0.25 M solution what volume is needed?

Ans.)

Molar mass of CH3OH$CH_{3}OH$

= (1 * 12) + (4 * 1) + (1 * 16)

= 32 g mol1$mol^{-1}$

= 0.032 kg mol1$mol^{-1}$

Molarity of the solution

= 0.793kgL10.032kgmol1$\frac{0.793\;kg\;L^{-1}}{0.032\;kg\;mol^{-1} }$

= 24.78 molL1$L^{-1}$

(From the definition of density)

M1V1=M2V2$M_{1}V_{1} = M_{2}V_{2}$

$∴$ (24.78 molL1$L^{-1}$) V1$V_{1}$ =  (2.5 L) (0.25 molL1$L^{-1}$)

V1$V_{1}$ = 0.0252 Litre

V1$V_{1}$ = 25.22 Millilitre

Q13. Pressure is defined as force per unit area of the surface. Pascal, the SI unit of pressure is as given below:

1 Pa = 1 N m2$m^{-2}$

Assume that mass of air at the sea level is 1034 gcm-2. Find out the pressure in Pascal.

Ans.

As per definition, pressure is force per unit area of the surface.

P = FA$\frac{F}{A}$

= 1034g×9.8ms2cm2×1kg1000g×(100)2cm21m2$\frac{1034\;g\;\times \;9.8\;ms^{-2}}{cm^{2}}\times \frac{1\;kg}{1000\;g}\times \frac{(100)^{2}\;cm^{2}}{1 m^{2}}$

= 1.01332 × 105$10^{5}$ kg m1s2$m^{-1} s^{-2}$

Now,

1 N = 1 kg ms2$s^{-2}$

Then,

1 Pa = 1 Nm2$Nm^{-2}$

= 1 kgm2$kgm^{-2}$s2$s^{-2}$

Pa   = 1 kgm1$kgm^{-1}$s2$s^{-2}$

$∴$ Pressure  (P) = 1.01332 × 105$10^{5}$ Pa

Q14. Write SI unit for mass. Also define mass.

Ans.

Si Unit: Kilogram (kg)

Mass:

“The mass equal to the mass of the international prototype of kilogram is known as mass.”

Q15. Match the prefixes with their multiples in the table given below:

 Prefixes Multiples (a) femto 10 (b) giga 10−15$10^{-15}$ (c) mega 10−6$10^{-6}$ (d) deca 109$10^{9}$ (e) micro 106$10^{6}$

Ans.

 Prefixes Multiples (a) femto 10−15$10^{-15}$ (b) giga 109$10^{9}$ (c) mega 106$10^{6}$ (d) deca 10 (e) micro 10−6$10^{-6}$

Q16. What are significant figures?

Ans.

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 the shows the uncertainty of the result is known as significant figures.”

Q17. A sample of drinking water was found to be highly contaminated with CHCl3$CHCl_{3}$, chloroform, which is carcinogenic. 15 ppm (by mass) was the level of contamination.

(a) Express in terms of percent by mass.

(b) Calculate the molality of chloroform in the given water sample.

Ans.

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

Mass percent of 15 ppm chloroform in H2O

= 15106×100$\frac{15}{10^{6}} \times 100$

= $\approx$ 1.5 ×103$10^{-3}$ %

(b) 100 gram of the sample is having 1.5 ×103$10^{-3}$g of CHCl3$CHCl_{3}$.

1000 gram of the sample is having 1.5 ×102$10^{-2}$g of CHCl3$CHCl_{3}$.

$∴$ Molality of CHCl3$CHCl_{3}$ in water

= 1.5×102gMolarmassofCHCl3$\frac{1.5 \;\times 10^{-2} \;g}{Molar \; mass \; of \; CHCl_{3}}$

Molar mass (CHCl3$CHCl_{3}$)

= 12 + 1 + 3 (35.5)

= 119.5 gram mol1$mol^{-1}$

Therefore, molality of CHCl3$CHCl_{3}$ I water

= 1.25 × 104$10^{-4}$ m

Q18. Express the given number in scientific notation:

(a) 0.0047

(b) 235,000

(c)8009

(d)700.0

(e) 5.0013

Ans.

(a) 0.0047= 4.7 ×103$10^{-3}$

(b) 235,000 = 2.35 ×105$10^{5}$

(c) 8009= 8.009 ×103$10^{3}$

(d) 700.0 = 7.000 ×102$10^{2}$

(e) 5.0013 = 5.0013

Q19. Find the number of significant figures in the numbers given belowt.

(a) 0.0027

(b) 209

(c)6005

(d)136,000

(e) 900.0

(f)2.0035

Ans.

(i) 0.0027: 2 significant numbers.

(ii) 209: 3 significant numbers.

(iii)6005: 4 significant numbers.

(iv)136,000:3 significant numbers.

(v) 900.0: 4 significant numbers.

(vi)2.0035: 5 significant numbers.

Q20. Round up the given numbers upto 3 significant numbers.

(a) 35.217

(b) 11.4108

(c)0.05577

(d)2806

Ans.

(a) The number after round up is: 35.2

(b) The number after round up is: 11.4

(c)The number after round up is: 0.0560

(d)The number after round up is: 2810

Q21. When dioxygen and dinitrogen react together, they form various compounds. The information is given below:

 Mass of dioxygen Mass of dinitrogen (i) 16 g 14 g (ii) 32 g 14 g (iii) 32 g 28 g (iv) 80 g 28 g

(1) In the data given above, which chemical combination law is obeyed? Also give the statement of the law.

(2) Convert the following:

(a) 1 km =_____ mm = _____ pm

(b) 1 mg = _____ kg = _____ ng

(c) 1 mL = _____ L =_____  dm3$dm^{ 3 }$

Ans.

(1) If we fix the mass of N2 at 28 g, the masses of N2 that will combine with the fixed mass of N2 are 32 gram, 64 gram, 32 gram and 80 gram.

The mass of O2 bear whole no. ratio of 1: 2: 2: 5. Therefore, the given information obeys the law of multiple proportions.

The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”

(2) Convert:

(a) 1 km = ____ mm = ____ pm

• 1 km = 1 km * 1000m1km$\frac{ 1000 \; m }{ 1 \; km }$ × 100cm1m$\frac{ 100 \; cm }{ 1 \; m }$ * 10mm1cm$\frac{ 10 \; mm }{ 1 \; cm }$

$∴$ 1 km = 106$10^{ 6 }$ mm

• 1 km = 1 km * 1000m1km$\frac{ 1000 \; m }{ 1 \; km }$ * 1pm1012m$\frac{1 \; pm}{10^{ -12 } \; m}$

$∴$ 1 km = 1015$10^{ 15 }$ pm

Therefore, 1 km = 106$10^{ 6 }$ mm = 1015$10^{ 15 }$ pm

(b) 1 mg = ____ kg = ____ ng

• 1 mg = 1 mg * 1g1000mg$\frac{ 1 \; g }{ 1000 \; mg }$ * 1kg1000g$\frac{ 1 \; kg }{ 1000 \; g }$

1 mg = 106$10^{ -6 }$ kg

• 1 mg = 1 mg * 1g1000mg$\frac{ 1 \; g }{ 1000 \; mg }$ * 1ng109g$\frac{ 1 \; ng }{ 10^{ -9 } \; g }$

1 mg = 106$10^{ 6 }$ ng

Therefore, 1 mg = 106$10^{ -6 }$ kg = 106$10^{ 6 }$ ng

(c) 1 mL = ____ L = ____ dm3$dm^{ 3 }$

• 1 mL = 1 mL * 1L1000mL$\frac{1 \; L}{1000 \; mL}$

1 mL = 103$10^{ -3 }$ L

• 1 mL = 1 cm3$cm^{ 3 }$ = 1 * 1dm×1dm×1dm10cm×10cm×10cmcm3$\frac{1 \; dm \; \times \; 1 \; dm \; \times \;1 \; dm }{10 \; cm \; \times \; 10 \; cm \; \times \; 10 \; cm } cm^{ 3 }$

1 mL = 103dm3$10^{ -3 } dm^{ 3 }$

Therefore, 1 mL = 103$10^{ -3 }$L = 103$10^{ -3 }$ dm3$dm^{ 3 }$

Q22. What is the distance covered by the light in 2 ns if the speed of light is 3 × 108$10^{ 8 }$ ms1$ms^{ -1 }$

Ans.

Time taken = 2 ns

= 2 × 109$10^{ -9 }$ s

Now,

Speed of light = 3 × 108$10^{ 8 }$ ms1$ms^{ -1 }$

So,

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

= (3 × 108$10^{ 8 }$)(2 × 109$10^{ -9 }$)

= 6 × 101$10^{ -1 }$ m

= 0.6 m

Q23. In the reaction given below:

X+Y2XY2$X \; + \; Y_{ 2 } \; \rightarrow \; XY_{ 2 }$

Find the limiting reagent if it is present in the  reactions given below:

(a) 2 mol X + 3 mol Y

(b) 100 atoms of X + 100 molecules of Y

(c) 300 atoms of X + 200 molecules of Y

(d) 2.5 mol X + 5 mol Y

(e) 5 mol X + 2.5 mol Y

Ans.

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 limiting the amt. of products formed.

(a) 2 mol X + 3 mol Y

1 mole of X reacts with 1 mole of Y. Similarly, 2 moles of X reacts with 2 moles of Y, so 1 mole of Y is unused. Hence, X is limiting agent.

(b) 100 atoms of X + 100 molecules of Y

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

(c) 300 atoms of X + 200 molecules of Y

1 atom of X reacts with 1 molecule of Y. Similarly, 200 atoms of X reacts with 200 molecules of Y, so 100 atoms of X are unused. Hence, Y is limiting agent.

(d) 2.5 mol X + 5 mol Y

1 mole of X reacts with 1 mole of Y. Similarly, 2.5 moles of X reacts with 2.5 moles of Y, so 2.5 mole of Y is unused. Hence, X is limiting agent.

(e) 5 mol X + 2.5 mol Y

1 mole of X reacts with 1 mole of Y. Similarly 2.5 moles of X reacts with 2 moles of Y, so 2.5 mole of X is unused. Hence, Y is limiting agent.

Q24. H2 and N2 react with each other to produce NH3 according to the given chemical equation

N2(g)+H2(g)2NH3(g)$N_{ 2 }\; (g) \; + \; H_{ 2 }\; (g) \; \rightarrow \; 2NH_{ 3 }\;(g)$

(a) What is the mass of NH3$NH_{ 3 }$ produced if 2×103$2 \; \times \;10^{ 3 }$ g N2 reacts with 1×103$1 \; \times \;10^{ 3 }$ g of H2?

(b) Will the reactants N2 or H2 remain unreacted?

(c) If any, then which one and give it’s mass.

Ans.

(a) Balance the given equation:

N2(g)+3H2(g)2NH3(g)$N_{ 2 }\;(g) \; + \; 3H_{ 2 } \;(g) \; \rightarrow \; 2NH_{ 3 }\;(g)$

Thus, 1 mole (28 g) of N2 reacts with 3 mole (6 g) of H2 to give 2 mole (34 g) of NH3$NH_{ 3 }$.

2×103$2 \; \times \;10^{ 3 }$ g of N2 will react with 6g28g×2×103$\frac{ 6 g }{ 28 g } \; \times \; 2 \; \times \; 10^{ 3 }$ g H2

2×103$2 \; \times \;10^{ 3 }$ g  of N2 will react with 428.6 g of H2.

Given:

Amt of H2 = 1×103$1 \; \times \;10^{ 3 }$

Therefore, N2$N_{ 2 }$ is limiting reagent.

28 g of N2$N_{ 2 }$ produces 34 g of NH3$NH_{ 3 }$

Therefore, mass of NH3$NH_{ 3 }$ produced by 2000 g of N2$N_{ 2 }$

= 34g28g×2000$\frac{ 34 \; g }{ 28 \; g } \; \times \; 2000$ g

(b) N2$N_{ 2 }$ is limiting reagent and H2$H_{ 2 }$ is the excess reagent. Therefore, H2$H_{ 2 }$ will not react.

(c) Mass of H2 unreacted

= 1×103$1 \; \times \;10^{ 3 }$ – 428.6 g

= 571.4 g

Q25. 0.50 mol Na2CO3$Na_{ 2 }CO_{ 3 }$ and 0.50 M Na2CO3$Na_{ 2 }CO_{ 3 }$ are different. How?

Ans.

Molar mass of Na2CO3$Na_{ 2 }CO_{ 3 }$:

= (2 × 23) + 12 + (3 × 16)

= 106 g mol1$mol^{ -1 }$

1 mole of Na2CO3$Na_{ 2 }CO_{ 3 }$ means 106 g of Na2CO3$Na_{ 2 }CO_{ 3 }$

Therefore, 0.5 mol of Na2CO3$Na_{ 2 }CO_{ 3 }$

= 106g1mol×0.5mol$\frac{ 106 \; g }{ 1 \; mol } \; \times \; 0.5 \; mol$ Na2CO3$Na_{ 2 }CO_{ 3 }$

= 53 g of Na2CO3$Na_{ 2 }CO_{ 3 }$

0.5 M of Na2CO3$Na_{ 2 }CO_{ 3 }$ = 0.5 mol/L Na2CO3$Na_{ 2 }CO_{ 3 }$

Hence, 0.5 mol of Na2CO3$Na_{ 2 }CO_{ 3 }$ is in 1 L of water or 53 g of Na2CO3$Na_{ 2 }CO_{ 3 }$ is in 1 L of water.

Q26. If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of vapour would be obtained?

Ans.

Reaction:

2H2(g)+O2(g)2H2O(g)$2H_{ 2 }\;(g) \; + \; O_{ 2 }\; (g) \; \rightarrow \; 2H_{ 2 }O\; (g)$

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

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

Q27. Convert the given quantities into basic units:

(i) 29.7 pm

(ii) 16.15 pm

(iii) 25366 mg

Ans.

(i) 29.7 pm

1 pm = 1012m$10^{ -12 } \; m$

29.7 pm = 29.7 × 1012m$10^{ -12 } \; m$

= 2.97 × 1011m$10^{ -11 } \; m$

(ii) 16.15 pm

1 pm = 1012m$10^{ -12 } \; m$

16.15 pm = 16.15 × 1012m$10^{ -12 } \; m$

= 1.615 × 1011m$10^{ -11 } \; m$

(iii) 25366 mg

1 mg = 103g$10^{ -3 } \; g$

25366 mg = 2.5366 × 101$10^{ -1 }$ × 103kg$10^{ -3 } \; kg$

25366 mg = 2.5366 × 102kg$10^{ -2 } \; kg$

Q28. Which of the given below have the largest no. of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl2$Cl_{ 2 }$ (g)

Ans.

(i) 1 g Au (s)

= 1197$\frac{ 1 }{ 197 }$ mol of Au (s)

= 6.022×1023197$\frac{ 6.022 \; \times \; 10^{ 23 } }{ 197 }$ atoms of Au (s)

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

(ii) 1 g Na (s)

= 123$\frac{ 1 }{ 23 }$ mol of Na (s)

= 6.022×102323$\frac{ 6.022 \; \times \; 10^{ 23 } }{ 23 }$ atoms of Na (s)

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

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

(iii) 1 g Li (s)

= 17$\frac{ 1 }{ 7 }$ mol of Li (s)

= 6.022×10237$\frac{ 6.022 \; \times \; 10^{ 23 } }{ 7 }$ atoms of Li (s)

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

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

(iv)1 g of Cl2$Cl_{ 2 }$ (g)

= 171$\frac{ 1 }{ 71 }$ mol of Cl2$Cl_{ 2 }$ (g)

(Molar mass of Cl2$Cl_{ 2 }$ molecule = 35.5 × 2 = 71 g mol1$mol^{ -1 }$)

= 6.022×102371$\frac{ 6.022 \; \times \; 10^{ 23 } }{ 71 }$ atoms of Cl2$Cl_{ 2 }$ (g)

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

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

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

Q29. What is the molarity of the solution of ethanol in water in which the mole fraction of ethanol is 0.040?

(Assume the density of water to be 1)

Ans.

Mole fraction of C2H5OH$C_{ 2 }H_{ 5 }OH$

= NumberofmolesofC2H5OHNumberofmolesofsolution$\frac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}$

0.040 = nC2H5OHnC2H5OH+nH2O$\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; n_{H_{ 2 }O}}$ ——(1)

No. of moles present in 1 L water:

nH2O=1000g18gmol1$n_{ H_{ 2 }O} \; = \; \frac{ 1000 \; g}{18 \; g \; mol^{ -1 }}$

nH2O$n_{ H_{ 2 }O}$ = 55.55 mol

Substituting the value of nH2O$n_{ H_{ 2 }O}$ in eq (1),

nC2H5OHnC2H5OH+55.55$\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; 55.55}$ = 0.040

nC2H5OH$n_{C_{ 2 }H_{ 5 }OH}$ = 0.040nC2H5OH$n_{C_{ 2 }H_{ 5 }OH}$ + (0.040)(55.55)

0.96nC2H5OH$n_{C_{ 2 }H_{ 5 }OH}$ = 2.222 mol

nC2H5OH$n_{C_{ 2 }H_{ 5 }OH}$ = 2.2220.96mol$\frac{ 2.222 }{ 0.96 } \; mol$

nC2H5OH$n_{C_{ 2 }H_{ 5 }OH}$ = 2.314 mol

Therefore, molarity of solution

= 2.314mol1L$\frac{ 2.314 \; mol }{ 1 \; L }$

= 2.314 M

Q30. Calculate the mass of 1 12 C $_{}^{ 12 }\textrm{ C }$ atom in g.

Ans.

1 mole of carbon atoms

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

= 12 g of carbon

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

= 12g6.022×1023$\frac{ 12 \; g }{ 6.022 \; \times \; 10^{ 23 }}$

= 1.993×1023g$1.993 \; \times \; 10^{ -23 } g$

Q31. How many significant numbers should be present in answer of the given calculations?

(i) 0.02856×298.15×0.1120.5785$\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }$

(ii) 5 × 5.365

(iii) 0.012 + 0.7864 + 0.0215

Ans.

(i) 0.02856×298.15×0.1120.5785$\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }$

Least precise no. of calculation = 0.112

Therefore, no. of significant numbers in the answer

= No. of significant numbers in the least precise no.

= 3

(ii) 5 × 5.365

Least precise no. of calculation = 5.365

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 5.365

= 4

(iii) 0.012 + 0.7864 + 0.0215

As the least no. of decimal place in each term is 4, the no. of significant numbers in the answer is also 4.

Q32. Calculate molar mass of Argon isotopes according to the data given in the table.

 Isotope Molar mass Abundance 36Ar$\, _{ 36 }\textrm{Ar}$ 35.96755 gmol−1$g \; mol^{ -1 }$ 0.337 % 38Ar$\, _{ 38 }\textrm{Ar}$ 37.96272 gmol−1$g \; mol^{ -1 }$ 0.063 % 40Ar$\, _{ 40 }\textrm{Ar}$ 39.9624 gmol−1$g \; mol^{ -1 }$ 99.600 %

Ans.

Molar mass of Argon:

= [(35.96755×0.337100)$( 35.96755 \; \times \; \frac{ 0.337 }{ 100 })$ + (37.96272×0.063100)$( 37.96272 \; \times \; \frac{ 0.063 }{ 100 })$ + (39.9624×99.600100)$( 39.9624 \; \times \; \frac{ 99.600 }{ 100 })$]

= [0.121 + 0.024 + 39.802] gmol1$g \; mol^{ -1 }$

= 39.947 gmol1$g \; mol^{ -1 }$

Q33. What is the number of atoms in the following compounds?

(i) 52 moles of Ar

(ii) 52 u of He

(iii) 52 g of He

Ans.

(i) 52 moles of Ar

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

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

= 3.131×1025$3.131 \; \times \; 10^{ 25 }$ 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 = 14$\frac{ 1 }{ 4 }$ atom of He

52 u of He = 524$\frac{ 52 }{ 4 }$ atom of He

= 13 atoms of He

(iii) 52 g of He

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

52 g of He = 6.023×1023×524$\frac{ 6.023 \; \times \; 10^{ 23 } \; \times \;52 }{ 4 }$ atoms of He

= 7.8286×1024$7.8286 \; \times \; 10^{ 24 }$ atoms of He

Q34. A welding fuel gas contains hydrogen and carbon. If we burn a small sample, we get 3.38 g of carbon dioxide and 0.69 g of water. A volume of 10 L (at STP) of this welding gas weighs 11.6 g.

Find:

(i) Empirical formula

(ii) Molar mass of the gas, and

(iii) Molecular formula

Ans.

(i) Empirical formula

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

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

= 12g44g×3.38g$\frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g$

= 0.9217 g

18 g of water contains 2 g of hydrogen

Therefore, 0.690 g of water will contain hydrogen

= 2g18g×0.690$\frac{ 2 \; g }{ 18 \; g } \; \times 0.690$

= 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

= 0.9217g0.9984g×100$\frac{ 0.9217 \; g }{ 0.9984 \; g } \; \times 100$

= 92.32 %

% of H in the compound

= 0.0767g0.9984g×100$\frac{ 0.0767 \; g }{ 0.9984 \; g } \; \times 100$

= 7.68 %

Moles of C in the compound,

= 92.3212.00$\frac{ 92.32 }{ 12.00 }$

= 7.69

Moles of H in the compound,

= 7.681$\frac{ 7.68 }{ 1 }$

= 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, and

Weight of 10 L of gas at STP = 11.6 g

Therefore, weight of 22.4 L of gas at STP

= 11.6g10L×22.4L$\frac{ 11.6 \; g }{ 10 \; L } \; \times \; 22.4 \; L$

= 25.984 g

$\approx$ 26 g

(iii) Molecular formula

Empirical formula mass:

CH = 12 + 1

= 13 g

n = MolarmassofgasEmpiricalformulamassofgas$\frac{ Molar \; mass \; of \; gas}{Empirical \; formula \; mass \; of \; gas}$

= 26g13g$\frac{ 26 \; g }{ 13 \; g}$

= 2

Therefore, molecular formula is (CH)n$(CH)_{ n }$ that is C2H2$C_{ 2 }H_{ 2 }$.

Q35. Calcium carbonate reacts with aqueous HCl and gives CaCl2$CaCl_{ 2 }$ and CO2$CO_{ 2 }$ according to the reaction:

CaCO3(s)+2HCl(aq)CaCl2(aq)+CO2(g)+H2O(l)$CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq) \; \rightarrow \; CaCl_{ 2 }\;(aq) \; + \; CO_{ 2 }\; (g) \; + \; H_{ 2 }O\; (l)$

Calculate the mass of CaCO3$CaCO_{ 3 }$ required to react completely with 25 mL of 0.75 M HCl?

Ans.

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 contins 27.375 g of HCl

Therefore, amt of HCl present in 25 mL of solution

= 27.375g1000mL×25mL$\frac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL$

= 0.6844 g

Given chemical reaction,

CaCO3(s)+2HCl(aq)CaCl2(aq)+CO2(g)+H2O(l)$CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq) \; \rightarrow \; CaCl_{ 2 }\;(aq) \; + \; CO_{ 2 }\; (g) \; + \; H_{ 2 }O\; (l)$

2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO3$CaCO_{ 3 }$ (100 g)

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

= 10071×0.6844g$\frac{ 100 }{ 71 } \; \times \; 0.6844 \; g$

= 0.9639 g

Q36. Chlorine is prepared by adding manganese dioxide with hydrochloric acid acc. to the reaction.

4HCl(aq)+MnO2(s)2H2O(l)+MnCl2(aq)+Cl2(g)$4 \; HCl\;(aq) \; + \; MnO_{ 2 }\;(s) \; \rightarrow \; 2 \; H_{ 2 }O\; (l) \; + \; MnCl_{ 2 }\; (aq) \; + \; Cl_{ 2 }\; (g)$

How many grams of HCl react with 5 g of manganese dioxide?

Ans.

1 mol of MnO2$MnO_{2}$ = 55 + 2 × 16 = 87 g

4 mol of HCl = 4 × 36.5 = 146 g

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

5 g of MnO2$MnO_{ 2 }$will react with:

= 146g87g×5g$\frac{146 \; g}{87 \; g} \; \times \; 5 \; g$ HCl

= 8.4 g HCl

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