Calculate the relative lowering of vapour pressure of solution containing 3 g urea in 50 g water. [ molar mass of urea $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ ]

Using Raoult's law, the relative lowering of vapor pressure equals the mole fraction of solute: $\frac{\Delta P}{P^0} = X_{\text{urea}}$.

Given 3 g urea (molar mass 60 g mol$^{-1}$) and 50 g water (molar mass 18 g mol$^{-1}$), the moles are:

$n_{\text{urea}} = \frac{3}{60} = 0.05$ mol

$n_{\text{water}} = \frac{50}{18} \approx 2.7778$ mol

The mole fraction of urea is:

$X_{\text{urea}} = \frac{0.05}{0.05 + 2.7778} = \frac{0.05}{2.8278} \approx 0.01768$

Rounding to three decimal places gives $\frac{\Delta P}{P^0} \approx 0.018$, which corresponds to option C.

Final answer: $\boxed{\text{C}}$