INTRODUCTION - Chapter 2 - Solutions - Exercise Solutions - Ncert Solutions class 12 - Chemistry

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Question 2.1:

Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Answer:

Mass percentage of C₆H₆ 

Mass percentage of CCl₄

Alternatively,

Mass percentage of CCl₄ = (100 − 15.28)%

= 84.72%

Question 2.2:

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

Answer:

Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

∴Mass of carbon tetrachloride = (100 − 30)g

= 70 g

Molar mass of benzene (C₆H₆) = (6 × 12 + 6 × 1) g mol⁻¹

= 78 g mol⁻¹

∴Number of moles of 

= 0.3846 mol

Molar mass of carbon tetrachloride (CCl₄) = 1 × 12 + 4 × 355

= 154 g mol⁻¹

∴Number of moles of CCl₄ 

= 0.4545 mol

Thus, the mole fraction of C₆H₆ is given as:

= 0.458

Question 2.3:

Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO₃)₂. 6H₂O in 4.3 L of solution (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

Answer:

Molarity is given by:

(a) Molar mass of Co (NO₃)₂.6H₂O = 59 + 2 (14 + 3 × 16) + 6 × 18

= 291 g mol⁻¹

∴Moles of Co (NO₃)₂.6H₂O

= 0.103 mol

Therefore, molarity 

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H₂SO₄ = 0.5 mol

∴Number of moles present in 30 mL of 0.5 M H₂SO₄ 

= 0.015 mol

Therefore, molarity

= 0.03 M

Question 2.4:

Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.

Answer:

Molar mass of urea (NH₂CONH₂) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16

= 60 g mol⁻¹

0.25 molar aqueous solution of urea means:

1000 g of water contains 0.25 mol = (0.25 × 60)g of urea

= 15 g of urea

That is,

(1000 + 15) g of solution contains 15 g of urea

Therefore, 2.5 kg (2500 g) of solution contains 

= 36.95 g

= 37 g of urea (approximately)

Hence, mass of urea required = 37 g

Note: There is a slight variation in this Answer and the one given in the NCERT textbook.

Question 2.5:

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.

Answer:

(a) Molar mass of KI = 39 + 127 = 166 g mol⁻¹

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 − 20) g of water = 80 g of water

Therefore, molality of the solution 

= 1.506 m

= 1.51 m (approximately)

(b) It is given that the density of the solution = 1.202 g mL⁻¹

∴Volume of 100 g solution 

= 83.19 mL

= 83.19 × 10⁻³ L

Therefore, molarity of the solution 

= 1.45 M

(c) Moles of KI 

Moles of water 

Therefore, mole fraction of KI 

= 0.0263