NCERT Chemistry Class 12 Chapter 2 Intext Solutions

**NCERT**

**Intext Questions**

**Chapter – Solutions**

**Que 1. ****Calculate the mass percentage of benzene (C _{6}H_{6}) and carbon tetrachloride (CCI_{4}) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride**.

**Ans 1.**Mass of solution = Mass of Benzene (C

_{6}H

_{6}) + Mass of Carbon Tetrachloride (CCl

_{4})

= 22 g +122 g = 144 g

Mass percentage of Benzene = 22/144 × 100 =15.28 %

Mass % of CCl

_{4}= 122/144 × 100 = 84.72 %

**Que 2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.**

**Ans 2.** Let the mass of Solution = 100 g

∴ Mass of Benzene in the solution = 30 g

∴ Mass of Carbon Tetrachloride = 100 – 30g = 70g

Molar mass of Benzene (C_{6}H_{6}) = 78 g mol^{-1}

Molar mass of CCl_{4} = 12 + 4 × 35.5 = 154 g mol^{-1}

∴ No. of moles of benzene** = **30/78 mol = = 0.3846 mol

Molar mass of CCl_{4} = 1 × 12 + 4 × 35.5 = 154 g/mol

Therefore, Number of moles of CCl_{4} = 70/154 = 0.4545 mol

Thus, the mole fraction of C_{6}H_{6} is given as

NCERT Chemistry Class 12 Chapter 2 Intext Solutions

**Que 3.** **Calculate the molarity of each of the following solutions**

**(a) 30 g of Co(NO _{3})_{2}6H_{2}O in 4·3 L of solution**

**(b) 30 mL of 0-5 M H**

_{2}SO_{4}diluted to 500 mL.**Ans 3.**We know that,

Molarity =

**(a)**Molar mass of Co(NO)

_{3}.6H

_{2}O = 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g/mol

Therefore, Moles of Co(NO)

_{3}.6H

_{2}O = 30/291 mol

= 0.103 mol

Therefore, Molarity = 0.1o3/4.3 mol/lt = 0.023 M

**(b)** Number of moles present in 1000 mL of 0.5 M H_{2}SO_{4} = 0.5 mol

Therefore, Number of moles present in 30 mL of 0.5 M H_{2}SO_{4} = 0.5 × (30/1000)mol

= 0.015 mol

Therefore, Molarity = 0.015/0.5 L = 0.03 M

**Que 4. ****Calculate the mass of urea (NH _{2}CONH_{2}) required in making 2.5 kg of 0.25 molal aqueous solution.**

**Ans 4.**Moles of urea = 0.25 mole

Mass of solvent (water) = 1 kg = 1000 g

Molar mass of urea (NH

_{2}CONH

_{2}) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g/mol

∴ Mass of urea in 1000 g of water = 0.25 mol × 60 g mol

^{-1}= 15 g

Total mass of solution = 1000 + 15 g = 1015 g

2.5 kg means 2500 g

1015 g of solution contain urea = 15 g

1 g of solution contains urea = 15/1015

Therefore, 2500 g of solution contains Urea = 15 × 2500/1015

= 37g

**Que 5. ****Calculate (a) molality (b) molarity and (c) mole fraction of Kl if the density of 20% (mass/mass) aqueous Kl is 1.202 g mL.**

**Ans 5. (a). ** **Calculation of molality of the solution**

Weight of KI in 100 g of the solution = 20 g

Weight of water in the solution = 100 – 20 = 80 g

Molar mass of KI = 39 + 127 = 166 g mol^{-1}

Therefore, the molality of the solution

= moles of KI**/**Mass of water in Kg = (20**/**166)**/** 0.08

= 1.506 m

= 1.51 m

**(b)** It is given that the density of the solution = 1.202 g mL^{-1}

Mass / Density = Volume of 100 g solution

100g / 1.202 gml^{-1}

= 83.19 mL

= 83.19 × 10^{-3} L

Therefore, molarity of the solution = (20/166) / 83.19 × 10^{-3} L

= 1.45 M

**(c)** Moles of KI

Moles of water = 80/18 = 4.44

Therefore, the mole fraction of KI = moles of KI/moles of KI + moles of water = 0.12/0.12 + 4.44

= 0.0263

NCERT Chemistry Class 12 Chapter 2 Intext Solutions

**Que 6. ****H _{2}S a toxic gas with a rotten egg-like smell is used for qualitative analysis. If the solubility of H_{2}S in the water at STP is 0.195 m, calculate Henry’s law constant.**

**Ans 6.**The solubility of H

_{2}S in the water at STP = 0.195 m

Mass of H

_{2}O = 1000 g

Number of moles of Water = = 55.56 mol

**Que 7. ****Henry’s law constant for CO _{2} in water is 1.67 × 10^{8} Pa at 298 K. Calculate the quantity of CO_{2} in 500 mL of soda water when packed under 2.5 atm CO_{2} pressure at 298K.**

**Ans 7.**

NCERT Chemistry Class 12 Chapter 2 Intext Solutions

**Que 8. ****The vapour pressures of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.**

**Ans 8. **P^{0}_{A} = 450 mm of Hg

P^{0}_{B} = 700 mm of Hg

Ptotal = 600 mm of Hg

According to Raoult’s law:

P_{A} = P^{0}_{A}X_{A} P_{B} = P^{0}_{B}X_{B}

We know X_{A} + X_{B} = 1

= X_{B} = 1 – X_{A}

Therefor P_{B} = P^{0}_{B }(1-X_{A})

Therefore, total pressure, P_{Total} = P_{A} + P_{B}

P_{Total} = P^{0}_{A}X_{A} + P^{0}_{B }(1-X_{A})

P_{Total} = P^{0}_{A}X_{A} + P^{0}_{B }– P^{0}_{B} X_{A}

P_{Total} = (P^{0}_{A }– P^{0}_{B})X_{A} + P^{0}_{B }

600 = (450 – 700) X_{A} + 700

250 X_{A }= 100

X_{A} = 100/250 = 0.4

Therefore, X_{B} = 1 – X_{A} = 1 – 0.4 = 0.6

P_{A} = P^{0}_{A}X_{A} = 450 × 0.4 = 180 mm of Hg

P_{B} = P^{0}_{B}X_{B} = 700 × 0.6 = 420 mm of Hg

Now, in the vapour phase: Mole fraction of liquid A = P_{A }/ P_{A} + P_{B}

= 180 / 180 + 420

= 180/600 = 0.30

And, mole fraction of liquid B = 1 – 0.30 = 0.70

**Que 9. The ****Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH _{2}CONH_{2}) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.**

**Ans 9.**

NCERT Chemistry Class 12 Chapter 2 Intext Solutions

**Que 10. The boiling**** point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? The Molal elevation constant for water is 0.52 K kg mol ^{-1}.**

**Ans 10.**

**Que 11. ****Calculate the mass of ascorbic acid (vitamin C, C _{6}H_{8}O_{6}) to be dissolved in 75 g acetic acid to lower its melting point by 1.5°C, KF = 3.9 K kg mol^{-1}.**

**Ans 11.**Mass of acetic acid (w

_{1}) = 75 g

Molar mass of ascorbic acid (C

_{6}H

_{8}O

_{6}), M

_{2}= 6 x 12 + 8 x 1 + 6 x 16 = 176 g mol

^{-1}

Hence, the amount of ascorbic acid needed to be dissolved is 5.08 g.

**Que 12. ****Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.**

**Ans 12. **Volume of water (V) = 450 mL = 0.45 L

Temperature (T) = 37 + 273 = 310 K

** NCERT Chemistry Class 12 Chapter 2 Intext Solutions**

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