Explanation

(a)
According to Raoult’s law relative lowering of vapour pressure $\propto$ mole fraction of solute
Thus, mole fraction of solute = 0.0125
Mole fraction of a solute is related to the molality by the following expression.
$\left( \frac{1}{X} - 1 \right) = \frac{1000}{m_{B} \times m}$
where, X = mole fraction of solute
$m_{B}$ = moleular weight of solvent
m = molality
$\left( \ \frac{1}{0.0125} - 1 \right) = \frac{1000}{18 \times m}$
m=$\frac{12.5}{(1 - 0.0125) \times 18}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{12.5}{17.775}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.70$