Question

if an ideal solutions made by mixing 2 moles of benzene( p°= 266 mm Hg) and 3 moles of another liquid (P ° = 236 mm Hg). the total vapour pressure of solution at the same temperature

Answer 1

mole fraction for benzene =2/2+3=2/5
mole fraction of another liquid is =3/3+2=3/5

total pressure =266×2/5+236×3/5
= 106.4 +141.6
=248 mm Hg

Answer 2

Answer: the total vapor pressure of solution at the same temperature is 248 mm Hg.

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.

$p_A=x_Ap_A^0$ and $p_B=x_BP_B^0$

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.

$p_{total}=p_A+p_B$

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

$x_{benzene}=\frac{\text {moles of benzene}}{\text {total moles}}=\frac{2}{5}=0.4$

$x_{liquid}=\frac{\text {moles of liquid}}{\text {total moles}}=\frac{3}{5}=0.6$

$p_{benzene}^0=266mmHg$

$p_{liquid}^0=236mmHg$

$p_{total}=0.4\times 266+0.6\times 236=248mmHg$