Question

What is the vapour pressure of a solution of glucose which has an osmotic pressure of 3 atmospheres at ${20^ \circ }C$ ? The vapour pressure of water at ${20^ \circ }C$ is $17.39mm$ . Consider the density of solution equal to that of solvent.
A) 12.35mm
B) 14.35mm
C) 16.35mm
D) 17.35mm

Answer 1

Osmotic Pressure and Relative lowering of Vapour pressure are both Colligative Properties. The osmotic pressure can be given as the minimum pressure required for the flow of Solvent through a semipermeable membrane. The osmotic pressure is denoted by the symbol $\pi$.

Complete answer:
The osmotic pressure is given by the formula: $\pi V = CRT$. Where, c is the concentration, V is the volume, T is the temperature. The formula can be modified as: $\pi V = \dfrac{n}{V}RT$
The information provided to us is: $\pi = 3atm$, $T = {20^ \circ }C = 20 + 273 = 293K$ , ${P_0} = 17.39mm$
${P_0}$ is the vapour pressure of pure solvent. To find the vapour pressure of Glucose, we’ll use the formula: $\dfrac{{\Delta P}}{{{P_0}}} = \dfrac{{\pi M}}{{dRT}}$ -- (1)
Where, $\pi$ is the osmotic pressure, d is the density, M is the molar mass and T is the temperature. The term $\dfrac{{\Delta P}}{{{P_0}}}$ is the relative lowering of Vapour pressure. The relative lowering of Vapour Pressure can also be given as: $\dfrac{{\Delta P}}{{{P_0}}} = \dfrac{{{P_0} - P}}{{{P_0}}}$-- (2)
Where, P is the vapour pressure of solution. Substituting the given values in equation (1) also, substituting (2) in (1) we get:
$\dfrac{{17.39 - P}}{{17.39}} = \dfrac{{3 \times 180}}{{1 \times 0.0821 \times 293}}$
$P = 17.35mm$
(Molar mass of Glucose is $180g/mol$, the density of water is $1g/c{m^3}$ and the value of R is $0.0821Latm/mol$ )

The correct answer is Option (D).

Note:
Consider the formula $\pi V = \dfrac{n}{V}RT$ . For a dilute solution the volume of solvent can be equal to that of the solvent. If N is the number of moles of the solvent, M is the molar mass of the solvent and d is the density, the volume can be given as: $V = \dfrac{{N \times M}}{d}$
From the above equation after simplification, we derived that $\dfrac{n}{N} = \dfrac{{\pi M}}{{\rho RT}}$ $\dfrac{n}{N} = \dfrac{{\pi M}}{{\rho RT}}$ . From the Raoult’s Law we know that $\dfrac{{{P_0} - P}}{{{P_0}}} = \dfrac{n}{N} = \dfrac{{\pi M}}{{dRT}}$ $\dfrac{{{P_0} - P}}{{{P_0}}} = \dfrac{n}{N} = \dfrac{{\pi M}}{{dRT}}$ .