Question

The vapour pressure of water at 20 is 17.5 mm Hg, if 18 g of glucose ${C_6}{H_{12}}{O_6}$ added to 178.2 g water at 20, the vapour pressure of the resulting solution will be:
A.17.675 mm Hg
B.15.750 mm Hg
C.16.500 mm Hg
D.17. 325 mm Hg

Answer 1

Hint: According to Raoult’s Law, the vapour pressure of the solution can be calculated by, ${P_{solution}} = {P^\circ }{X_{solvent}}$
Which means, ${P_{solution}} = $ Vapour pressure of its pure component $ \times $ Mole fraction in solution.
Where, ${P_{solution}}$ is the vapour pressure of the solution, ${X_{solvent}}$ is the mole fraction of the solvent and ${P^\circ }$ is the vapour pressure of solvent.

Complete step by step answer:
When the molecules leaving the liquid and entering into the gaseous phase and the molecules leaving the gaseous phase and entering the liquid phase are in the equilibrium. This point of pressure is called Vapour Pressure of liquid.
The vapour pressure of water is the pressure at which the gas phase is in the equilibrium with the liquid phase.
Vapour pressure of water $({P^\circ })$ at ${20^\circ }C = 17.5$ mm Hg
Weight of glucose ${C_6}{H_{12}}{O_6}$ added $({W_8}) = 18g$
Weight of water $({W_A}) = 178.2g$
From Raoult’s Law, ${P_{solution}} = {P^\circ }{X_{solvent}}$
Which means, ${P_{solution}} = $ Vapour pressure of its pure component $ \times $ Mole fraction in solution
 ${X_{solvent}} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}$ where, ${n_A} = \dfrac{{{W_A}}}{{{M_A}}}$ and ${n_B} = \dfrac{{{W_B}}}{{{M_B}}}$
 $ \Rightarrow {X_{solvent}} = \dfrac{{\dfrac{{178.2}}{{18}}}}{{\dfrac{{18}}{{180}} + \dfrac{{178.2}}{{18}}}}$
 $ \Rightarrow {X_{solvent}} = \dfrac{{99}}{{9.94}}$
 $ \Rightarrow {X_{solvent}} = 0.99$
Now, ${P_{solution}} = {P^\circ }{X_{solvent}}$
 ${P_{solution}} = 17.5 \times 0.99$ = 17.325 mm Hg
So, the vapour pressure of the resulting solution is 17.325 mm Hg.

Therefore, the correct answer is option (D)

Note: The vapour pressure of a liquid varies with its temperature. As the temperature of the liquid increases, its kinetic energy of the molecule also increases which will thereby increase the vapour pressure. A substance with high vapour pressure at a normal temperature is often considered as volatile.