Question

At 25°c vapour pressure of volatile liquid A in pure state is 60 torr to 4 mole of A , x moles of B volatile liquid is added at 25°c the vapour pressure of resulting solution was found to be 72 torr and mole fraction of A in vapour phase is 1/3 , the value of x is​

Answer 1

Answer:

look the file and get the answer

Answer 2

Answer:

The value of $x$ is equal to $8$ moles.

Explanation:

Raoult's law states that the vapour pressure of a solvent above a solution is same as the vapour pressure of the natural solvent on the identical temperature scaled by the mole fraction of the solvent present:

$P_{\text {solution }}=X_{\text {solvent }} P_{\text {solvent }}$

$P_{\text {solution }}=\text { vapour pressure of the solution }$

$X_{\text {solvent }}=\text { mole fraction of the solvent }$

$P_{\text {solvent }}=\text { vapour pressure of the pure solvent }$

Initially, only A was present,

So, Pressure $=60$ torr

After adding B, $20 \%$ was reduced

Therefore,

$P_{Total} =48$ torr

$P_{\text {Total }}=P_{A}^{0}+X_{A}\left(P_{B}^{0}-P_{A}^{0}\right)$

$\begin{gathered}48=42+\frac{4}{4+x}(60-42) \\6(4+x) =4(18) \\x=8 \text { moles }\end{gathered}$

Therefore, the required value of $x$ is 8 moles.