Question

The vapour pressure of 98%aqueous solution of a is 700 mm of hg at 373 k.Thus the molecular weight of a is

Answer 1

Answer: The molecular weight of a is 220 g/mol.

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  (water) = 760 mmHg

$p_s$ = vapor pressure of solution  = 700 mmHg

Given : 98 g of a is dissolved in 100 g of the solvent (water)

$w_2$ = mass of solute (a) = 98 g

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

$M_1$ = molar mass of solvent (water) = 18 g/mole

$M_2$ = molar mass of solute (a) = ? g/mol

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

$\frac{760-700}{760}=\frac{98\times 18}{100\times M_2}$

$M_2=220g/mol$

Therefore, the molecular weight of a is 220 g/mol