At 25°C the vapour pressure of benzene, C6H6 (78 g/mole), is 93.2 Torr and that of toluene, C7H8 (92 g/mol), is 28.2 torr. A solution of 1.0 mole of C6H6 and 1.0 mol of C7H8 is prepared. Calculate the mole fraction of C6H6 in the vapour above this solution (assume the solution is ideal)

Question

At 25°C the vapour pressure of benzene, C6H6 (78 g/mole), is 93.2 Torr and that of toluene, C7H8 (92 g/mol), is 28.2 torr. A solution of 1.0 mole of C6H6 and 1.0 mol of C7H8 is prepared. Calculate the mole fraction of C6H6 in the vapour above this solution (assume the solution is ideal) (A) 0.607 (B) 0.768 (C) 0.232 (D) 0.393

Tardigrade Admin · Accepted Answer

YB=(PBo× XB/P) YB=(93.2×0.5/P) YT=(28.2×0.5/P) ∴ (YB/YT)=3.3 YB+YT=1 On solving YB=0.768

$0.607$

$0.768$

$0.232$

$0.393$

$Y_{B} = \frac{P _{B}^{o} \times X _{B}}{P}$
$Y_{B} = \frac{93.2 \times 0.5}{P}$
$Y_{T} = \frac{28.2 \times 0.5}{P}$
$∴ \frac{Y _{B}}{Y _{T}} = 3.3$ & $Y_{B} + Y_{T} = 1$
On solving $Y_{B} = 0.768$

0.607

0.768

0.232

0.393

YB​=PPBo​×XB​​ YB​=P93.2×0.5​ YT​=P28.2×0.5​ ∴YT​YB​​=3.3 & YB​+YT​=1 On solving YB​=0.768

$0.607$

$0.768$

$0.232$

$0.393$

$Y_{B} = \frac{P _{B}^{o} \times X _{B}}{P}$
$Y_{B} = \frac{93.2 \times 0.5}{P}$
$Y_{T} = \frac{28.2 \times 0.5}{P}$
$∴ \frac{Y _{B}}{Y _{T}} = 3.3$ & $Y_{B} + Y_{T} = 1$
On solving $Y_{B} = 0.768$