Calculate the vapour pressure of solution if relative lowering of vapour pressure and vapour pressure of pure solvent are 0.018 and 18 mm Hg respectively at 300 K .

Vapor pressure lowering is a colligative property described by Raoult's law for non-volatile solutes:

$\frac{P^0 - P_s}{P^0} = \text{relative lowering}$

Given a relative lowering of $0.018$ and a pure solvent vapor pressure $P^0 = 18\text{ mm Hg}$, substitute into the equation:

$\frac{18 - P_s}{18} = 0.018$

Solving for $P_s$:

$18 - P_s = 0.018 \times 18$

$18 - P_s = 0.324$

$P_s = 18 - 0.324 = 17.676\text{ mm Hg}$

Rounded to two decimal places, $P_s = 17.68\text{ mm Hg}$, which corresponds to option C.

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