Question

The vapour pressure of pure water at ${26^0}$ C is 25.21 torr. What is the vapour pressure of a solution which contains 20.0 g of glucose (${C_6}{H_{12}}{O_6}$), in 70 g water?
(A) 24.5
(B) 24.8
(C) 25.3
(D) 25.5

Answer 1

Hint: When a solute is added to a pure solvent, the vapour pressure of the pure solvent is decreased and hence, the vapour pressure of the solution is always less than that of the pure solvent. The decrement of the vapour pressure can be determined from Raoult's law.

Complete step by step answer
In order to find out the answer of this question, let’s first mention the concept of lowering of vapour pressure of a solvent.
When a solute is added to the solvent to prepare a solution, the vapour pressure of the solution is always less than the vapour pressure of the pure solvent.
Mathematically, if ${P_0}$ is the vapour pressure of the pure solvent, and ${P_A}$ is the vapour pressure of the solution after adding a solute ‘A’ to the solvent, then we can write:
${P_A} < {P_0}$
Now, according to Raoult's law for lowering vapour pressure, the relative lowering of vapour pressure is equal to the mole fraction of the solute.
Mathematically, if ${x_A}$ is mole fraction of the solute ‘A’, then we can write:
$\dfrac{{{P_0} - {P_A}}}{{{P_0}}} = {x_A}$ …(1)
Here, the term is called relative lowering of the vapour pressure.
We will use the equation (1) to solve this problem.
In this contrast, let us mention how to calculate the number of moles of a solvent or solute.
If weight of a component is ‘W’ and the molar mass of the component is ‘M’, then, number of moles of that component will be:
$n = \dfrac{W}{M}$
Now,
We have,
Weight of glucose: 20.0 g
Molar mass of glucose: 180
Hence, number of moles of glucose (${n_{glu\cos e}}$): $\dfrac{{20}}{{180}} = 0.111$
Again,
Weight of water: 70 g
Molar mass of water: 18
Hence, number of moles of water (${n_{water}}$): $\dfrac{{70}}{{18}} = 3.89$

Now,
Mole fraction of the solute (${x_A}$) is given by:
${x_A} = \dfrac{{{n_{glu\cos e}}}}{{{n_{glu\cos e}} + {n_{water}}}}$
Hence, we will have,
${x_A} = \dfrac{{0.111}}{{0.111 + 3.89}}$
${x_A} = 0.0277$
Thus, mole fraction of the solute glucose (${x_A}$) is 0.0277.
Now,
Vapour pressure of the pure water (${P_0}$) is 25.21 torr. Thus, if we use equation (1), then we can write:
$\dfrac{{25.21 - {P_A}}}{{25.21}} = 0.0277$
Or,
${P_A} = 25.21 - (0.0277 \times 25.21)$
Or,
${P_A} = 24.5$
Hence, the vapour pressure of the solution is 24.5 torr.
Hence, option A is the correct answer to this question.

Note: Students should not get confused in two things:
- Moles and mole fractions: moles are used to measure the amount of substance present in a system. Whereas, mole fraction is the ratio of moles of a particular component in a system to the summation of moles of all the components.
- Lowering of vapour pressure and relative lowering vapour pressure: lowering of vapour pressure implies the decrement of vapour pressure of a solution from the solvent, that is ${P_0} - {P_A}$. Whereas, relative lowering of vapour pressure means lowering of vapour pressure relative to the vapour pressure of pure solvent, that is $\dfrac{{{P_0} - {P_A}}}{{{P_0}}}$.