Question

The vapour pressures of pure liquids 'A' and 'B' are $300$ and $800\text{torr}$ respectively at $25{}^{\circ }\text{C}$. When these two liquids are mixed at this temperature to form a solution in which mole fraction of 'B' is $0.92$, then the total vapour pressure is observed to be $0.95\text{atm}$. Which of the following is true for this solution?

• A. $\Delta V_{\text{mix}}>0$
• B. $\Delta H_{\text{mix}}<0$
• C. $\Delta V_{\text{mix}}=0$
• D. $\Delta H_{\text{mix}}=0$

Answer 1

The correct option is B $\Delta H_{\text{mix}}<0$

Given, mole fraction of 'B' is $0.92$
Thus mole fraction of 'A' would be $0.08$
According to Raoult's law.
$P_T=(0.08\times 300+0.92\times 800)\text{torr}$
$=(24+736)\text{torr}$
$=760$ torr = $1\text{atm}$
$P_{\text{experimental}}=0.95\text{atm}<1\text{atm}$
Hence the solution shows negative deviation.
So, $\Delta H_{\text{mix}}<0$ and $\Delta V_{\text{mix}}<0$