Question

At $300K$, the vapour pressure of a solution containing $1$ mole of $n-hexaneand3$ moles of $n-heptaneis550mmofHg$. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by $10mmofHg$. What is the vapour pressure in $mmHg$ of n-heptane in its pure state ________?

• A. 600
• B. 600.0
• C. 600.00

Answer 1

Answer: (A), (B), (C)

If $X_1$ and $P_1^{\circ }$ are the mole fraction and vapour pressure of n -hexane in solution and $X_{2}and{P}_2^{\circ }$ are the mole fraction and vapour pressure of n- heptane in solution then

$550=X_1{P}_1^{\circ }+X_2{P}_2^{\circ }$

$X_1=\frac{1}{4};X_2=\frac{3}{4}$

$=\frac{P_1^{\circ }}{4}+\frac{3P_2^{\circ }}{4}\Rightarrow P_1^{\circ }+3P_2^{\circ }=2200...\left(1\right)$

On addition of 1 more mole of n - heptane

$X_1^{\prime }=\frac{1}{5};X_2^{\prime }=\frac{4}{5}$

$560=X_1^{\prime }{P}_1^{\circ }+X_2^{\prime }{P}_2^{\circ }$

$=\frac{P_1^{\circ }}{5}+\frac{4P_2^{\circ }}{5}\Rightarrow P_1^{\circ }+4P_2^{\circ }=2800...\left(2\right)$

From (1) and (2), we get,
$P_2^{\circ }=600mmHg$