Question

23. The vapour pressure of pure benzene at 298
K is 97.0 mm Hg. The vapour pressure of
solution containing 10.0 g of naphthalene
(mol. wt.128) in 100 g of benzene at 398 K
will be
A. 79.5 mm Hg B. 91.43 mm Hg
C. 81.45 mm Hg D. 66.5 mm Hg​

Answer 1

Answer: B. 91.43 mm Hg

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent  (benzene) = 97.0 mm Hg

$p_s$ = vapor pressure of solution = ?

$w_2$ = mass of solute  (napthalene) = 10.0 g

$w_1$ = mass of solvent  (benzene) = 100 g

$M_1$ = molar mass of solvent (benzene) = 78 g/mole

$M_2$ = molar mass of solute (nathalene) = 128 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{97-p_s}{97}=\frac{10\times 78}{100\times 128}$

$p_s=91.43 mm Hg$

Therefore, the vapor pressure of solution is, 91.43 mmHg