Question

| 79. 1.5 mole of n-hexane is mixed with 0.5 mole of
n-pentane to form an ideal solution at 25°C. If the
vapour pressure of n-hexane and n-pentane at
25°C are 150 mm and 200 mm of Hg respectively
then the vapour pressure of the resulting solution
in mm of Hg will be
(1) 215.5
(2) 250
$3) 162.5
(4) 175​

Answer 1

Answer: 3) 162.5 mmHg

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$

$p_{total}=x_Ap_A^0+x_BP_B^0$

$x_{n-hexane}=\frac{\text {moles of n-hexane}}{\text {moles of -hexane+moles of n-pentane}}=\frac{1.5}{1.5+0.5}=0.75$  

$x_{n-pentane}=\frac{\text {moles of n-pentane}}{\text {moles of n-hexane+moles of n-pentane}}=\frac{0.5}{1.5+0.5}=0.25$

$p_{n-hexane}^0=150mmHg$

$p_{n-pentane}^0=200mmHg$

$p_{total}=0.75\times 150+0.25\times 200=162.5mmHg$

The total vapor pressure above the solution at 25.0∘C is 162.5 mm Hg.