Question

The vapour pressure of benzene at 90°c is 1020 torr. A solution of 5g of solute in 58.5g of benzene has vapour pressure 990 torr. The molar mass of solute is solution

Answer 1

Answer : The molar mass of the solute is, 226.75 g/mole

Solution :

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,

$\frac{p^o-p_s}{p^o}=\frac{w_2\times M_1}{w_1\times M_2}$

where,

$p^o$ = vapor pressure of pure solvent  (benzene) = 1020 torr

$p_s$ = vapor pressure of solution  = 990 torr

$w_2$ = mass of solute  = 5 g

$w_1$ = mass of solvent  (benzene) = 58.5 g

$M_1$ = molar mass of solvent = 78 g/mole

$M_2$ = molar mass of solute = ?

Now put all the given values in this formula ,we get the molar mass of solute.

$\frac{1020-990}{1020}=\frac{5\times 78}{58.5\times M_2}$

$M_2=226.75g/mole$

Therefore, the molar mass of the solute is, 226.75 g/mole