The boiling point of benzene is 353 . 23 K . When 1 . 80 g of non-volatile solute was dissolved in 90 g of benzene, the boiling point of is raised to 354 . 11 K . Calculate the molar mass of the solute. ( K b for benzene is 2 . 53 K Kg mol - 1 )

Given, The boiling point of pure benzene = 353 . 23 K The boiling point of the solution = 354 . 11 K The weight of the benzene W A = 90 g The weight of the solute W B = 1 . 80 g Molal boiling point elevation constant K b = 2 . 53 K Kg mol - 1 So, the elevation in boiling point ∆ T b = 354 . 11 - 353 . 23 = 0 . 88 K If the molar mass of the solute is M B , then ⇒ M B = K b × W B × 1000 ∆ T b × W A ⇒ M B = 2 . 53 × 1 . 80 × 1000 0 . 88 × 90 = 57 . 5 g mol - 1 The molar mass of the solute is 57 . 5 g mol - 1 .

E M B I B E

All In One ISC Chemistry Class 12>Chapter 2 - Solutions>Chapter Practice>Q 45

MEDIUM

12th ICSE

IMPORTANT

Earn 100

At what partial pressure oxygen will have a solubility of $0.05 g L^{- 1}$ in water at $293 K$ ? (Henry's constant for oxygen in water at $293 K$ is $34.86 K bar$ ). Assume that the density of the solution to be same as that of the solvent.

Important Questions on Solutions

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 2 - Solutions>Chapter Practice>Q 46

The solubility of pure nitrogen gas at

25 ° C

and

1

atm is

6.8 \times 10^{- 4} mol L^{- 1}

. What is the concentration of nitrogen dissolved in the water under atmospheric conditions? The partial pressure of nitrogen gas in the atmosphere is