Question

On mixing, heptane and octane form an ideal solution. At 373 $K$, the vapour pressures of the two liquid components (heptane and octane) are 105 $kPa$ and 45 $kPa$ respectively. Vapour pressure of the solution obtained by mixing 25.0 gm of heptane and 35 gm of octane will be:(molar mass of heptane $=100gmo{1}^{-1}$ and of octane $=114gmo{1}^{-1}$)

• A. $72.0kPa$
• B. $36.1kPa$
• C. $96.2kPa$
• D. $144.5kPa$

Answer 1

The correct option is A $72.0kPa$

Mole fraction of Heptane $=\frac{25/100}{\frac{25}{100}+\frac{35}{114}}=\frac{0.25}{0.557}=0.45$

$X_{Heptane}=0.45$

$\therefore$ Mole fraction of octane $=1-0.45=0.55=X_{octane}$

Total pressure $=\sum X_i{P}_i^0$

$P_t=(105\times 0.45)+(45\times 0.55)kPa$

$P_t=72.0kPa$