CBSE Class 12 · 2013 · SolutionsLong Answer

WhatsApp Email

(a) State Raoult's law for a solution containing volatile components. How does Raoult's law become a special case of Henry's law?
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (K

_{f}

for benzene = 5.12 K kg mol

^{- 1}

)

Solution

(a) Raoult's law: in a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction,

p_{i} = x_{i} p_{i}^{\circ}

.
Henry's law for a volatile solute reads

p = K_{H} x

. Both state that the vapour pressure is proportional to the mole fraction; when the proportionality constant

K_{H}

equals the pure component's vapour pressure

p^{\circ}

, Henry's law reduces to Raoult's law — so Raoult's law is a special case of Henry's law.
(b) Molality:

m = \frac{Δ T _{f}}{K _{f}} = \frac{0.40}{5.12} = 0.078125 mol kg^{- 1}

Moles in 50 g benzene =

0.078125 \times 0.050 = 3.906 \times 1 0^{- 3}

mol.
Molar mass:

M = \frac{1.00}{3.906 \times 1 0 ^{- 3}}

M = 256 g mol^{- 1}

CBSE Class 12 Board · 2013 · Subjective

Practise more: all CBSE board PYQs on Solutions →

CBSE Class 12 · 2013 · SolutionsLong Answer

Solution

More board questions on Solutions