Benzene and napthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm of Hg and 32.06 mm of Hg respectively. What will be the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene?

Question

Benzene and napthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm of Hg and 32.06 mm of Hg respectively. What will be the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene? (A) 0.0675 (B) 0.675 (C) 0.35 (D) 0.5

Tardigrade Admin · Accepted Answer

(P ° A XA/P ° B XB)=(XA'/XB') (50.71((80/78))/32.06((100)128))=( mathrmX mathrmA prime/1- mathrmX mathrmA prime) XA'=0.675

0.0675

0.675

0.35

0.5

PB∘​XB​PA∘​XA​​=XB′​XA′​​ 32.06(128100​)50.71(7880​)​=1−XA​′XA′​​ XA′​=0.675

$\frac{P _{A}^{\circ} X _{A}}{P _{B}^{\circ} X _{B}} = \frac{X _{A}^{^{'}}}{X _{B}^{^{'}}}$
$\frac{50.71 ( \frac{80}{78} )}{32.06 ( \frac{100}{128} )} = \frac{X _{A}^{'}}{1 - X _{A} ^{'}}$
$X_{A}^{^{'}} = 0.675$