(a) State Raoult's law for a solution containing volatile components. How does Raoult's law become a special cause 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 = $5.12 K k g m o l^{- 1})$

Open in App

Solution

(a) Raoult's law : In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state. For two components A and B
$p_{A} = p_{A}^{\circ} x_{A} p_{B} = p_{B}^{\circ} x_{B}$
Raoult's law as a special case of Henry's law :
According to Raoult's law, for any volatile component of the solution.
$p_{A} = p_{A}^{\circ} \times x_{A}$
i.e., vapour pressure of the volatile component is directly proportional to the mole fraction of that component in the solution.
Now, if gas is the solute and liquid is the solvent, then according to Henry's law
$p_{A} = k_{H} \times x_{A}$
i.e., partial pressure of the volatile component (gas) is directly proportional to the mole fraction of that component (gas) in the solution.
Thus, Raoult's law and Henry's law become identical except that their proportionality constants are different.
(b) $W_{2} = 1.00 g$
$W_{1} = 50 g Δ T_{f} = 0.40 J K_{f} = 5.12 K k g m o l^{- 1}$
$M_{2} = \frac{1000 \times k_{f} \times W_{2}}{W_{1} \times Δ T_{f}}$
$= \frac{1000 \times 5.12 \times 1.00}{50 \times 0.40} = 256 g m o l - 1$

Suggest Corrections

Similar questions

1.00 g

250 g m o l^{- 1}

51.2 g

K_{f}

5.12 K k g m o l^{- 1}

0.850 b a r

0.5 g

39.0 g

(m o l a r m a s s 78 g m o l^{- 1})

0.845 b a r

(K_{f})

{5.12 K kg mol}^{- 1}

0.01 m

Join BYJU'S Learning Program