Question

1.4g of acetone dissolved in 100 g of benzene gave the solution which freeze at 271.12к. Pure benzere
freezes at 278.4k. 2.8 g of solid (A) dissolved in 100g of benzene gave solution which freezes
277.76 К Calculate the molar mass of A​

Answer 1

Given:

w1 = 1.4 gm

W1 = 100 gm

T1 = 271.12 K

T = 278.4 K

w2 = 2.8 gm

W2 = 100 gm

T2 = 277.76 K

To Find:

The molar mass of compound A.

Calculation:

- M.wt of Acetone, M.wt1 = 58 gm

- Depression in freezing point of acetone solution:

ΔTf = Kf × m1

⇒ (T - T1) = Kf × (w1 × 1000)/(M.wt1 × W1)

⇒ (278.4 - 271.12) = Kf × (1.4 × 1000)/(58 × 100)

⇒ 7.28 = Kf × 14/58

⇒ Kf = 30.16

- Depression in freezing point of the solution of A:

ΔTf = Kf × m2

⇒ (T - T2) = Kf × (w2 × 1000)/(M.wt × W2)

⇒ (278.4 - 277.76) = 30.16 × (2.8 × 1000)/(M.wt × 100)

⇒ 0.64 = 30.16 × 28/M.wt

⇒ M.wt = 844.48/0.64

⇒ M.wt = 1319.5 gm

- So, the molar mass of A​ is 1319.5 gm.