18 g. of a non-volatile solute is dissolved in 900 g of water such that the relative lowering of vapor pressure is 0.0019. The molecular weight of the solute is

18g of a non-volatile solute dissolved in 900g of water produces the relative lowering of the vapor pressure of 0.1. The molecular mass of the solute is 189.47.

 $\frac{\Delta \mathrm{P}}{{\mathrm{P}}^0}= 0.0019$

$\frac{\Delta \mathrm{P}}{{\mathrm{P}}^0}$ = $\frac{{\mathrm{W}}_2{\mathrm{M}}_1}{{ \mathrm{W}}_1{\mathrm{M}}_2}$

$0.0019$=$\frac{18 \mathrm{g}\times 18 \mathrm{g}/\mathrm{mol}}{900 \mathrm{g}\times {\mathrm{M}}_2}$

${\mathrm{M}}_2$ = $189.47 \mathrm{g}/\mathrm{mol}$

Note: ${\mathrm{W}}_1$ and ${\mathrm{W}}_2$ are weights of water and solute respectively.
${\mathrm{M}}_1$ and ${\mathrm{M}}_2$ are molecular weights of water and solute respectively.

Hence, the correct answer is option C).