An ideal binary liquid solution of $A$ and $B$ has a vapour pressure of $900 mm$ of $H g$ . It is distilled at constant temparature till $\frac{2}{3}$ of the original amount is distilled. If the mole fraction of $^{'} A^{'}$ in residue and mole fraction of $B$ in condensate are $0.3$ and $0.7$ respectively and vapour pressure of residue is $860 mm$ of Hg then idenetify the correct options.

The initial mixture taken should be an equimolar mixture of

A

and

B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

The first vapours formed will have more moles of

A

as compared to moles of

B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P_{A} = 1000 mm

of Hg

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

The vapour pressure above the condenstate will be

920

mm of Hg at same temparature

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D The vapour pressure above the condenstate will be $920$ mm of Hg at same temparature
Let the mole of $A$ & $B$ be $^{'} a^{'}$ & $^{'} b^{'}$ respectively in the initial mixture.
Initial moles of $A$ in the solution = Final moles of $A$ in the residue +Moles of $A$ in condensate
$a = 0.3 \times \frac{(a + b)}{3} + 0.6 \times \frac{2 (a + b)}{3}$
$\Rightarrow a = 0.1 a + 0.1 b + 0.4 a + 0.4 b$
$\Rightarrow a = 0.1 a + 0.1 b + 0.4 a + 0.4 b$
$\Rightarrow a = 0.5 a + 0.5 b$
$\Rightarrow 0.5 a = 0.5 b$
$\Rightarrow a = b$
Moles of $A$ and $B$ in the initial mixture are equal
$P = P_{A}^{o} χ_{A} + P_{b}^{o} χ_{B}$
$\Rightarrow 900 = P_{A}^{o} \frac{1}{2} + P_{B}^{o} \frac{1}{2}$
$\Rightarrow 1800 = P_{A}^{o} + P_{B}^{o} \cdot \cdot \cdot (1)$
In residue mole fraction for $A$ and $B$ be $χ_{A}^{/} = 0.3$ , $χ_{B}^{/} = 0.7$ respectively.
$P^{/} = P_{A}^{o} χ_{A}^{/} + P_{B}^{o} χ_{B}^{/}$
$\Rightarrow 860 = P_{A}^{o} (0.3) + P_{B}^{o} (0.7)$
$\Rightarrow 860 = 0.3 P_{A}^{o} + 0.7 P_{B}^{o} \cdot \cdot \cdot (2)$
On solving $(1)$ and $(2)$ we get,
$P_{A}^{o} = 1000 nm_, P_{B}^{o} = 800 nm$
In condensate mole fractions are, $X_{A}^{''} = 0.6$ , $X_{B}^{''} = 0.4$
Therefore, $P^{''} = P_{A}^{o} X_{A}^{''} + P_{A}^{o} X_{A}^{''} +$
$P^{''} = 1000 \times 0.6 + 800 \times 0.4 = 920 m$
In condensate mole fraction of $A$ is more than $B$ , $X_{A}^{''} > X_{B}^{''}$
Therefore, the first vapours formed will have more moles of $A$ as compared to moles of $B$

Suggest Corrections

Similar questions

3 moles of A ( $P_{A}^{0}$ =200mm of Hg) and 2 moles of B( $P_{B}^{0}$ =300mm of Hg) are mixed at a fixed temperature. The vapours above the solution are collected and condensed. Than:

Q. In a binary mixture of two ideal volatile liquids A and B, calculate the mole fraction of component A in vapour phase if:
Vapour pressure of A in pure state is

500 m m H g

Vapour pressure of B in pure state is

700 m m H g

Total pressure

(P_{T})

of the solution is

600 m m H g

Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapour pressure of 700 torr at $80^{\circ}$ C. It is distilled till $\frac{2}{3}$ rd of the solution (mole basis) is collected as condensate. The composition of the condensate is $X_{A}^{'} = 0.75$ and that of the residue is $X_{A}^{''} = 0.30$ . If the vapour pressure of the residue at $80^{\circ}$ C is 600 torr, which of the following option(s) is/are correct.