Question

If $18g$glucose is added to $178.2g$ of water the vapour pressure of water for this aqueous solution at ${100^\circ }C$ is:
A. $704torr$
B. $759torr$
C. $7.6torr$
D. None of these

Answer 1

Hint: Glucose is an example of carbohydrate. Vapour pressure is observed in case of solutions and also in pure solvents. The relation between number of moles of solute and vapour pressure of solvent is given by Raoult’s law.

Complete step by step answer:
Definition: Any liquid at given temperature can vaporize and at equilibrium the pressure exerted by the vapors of the liquid over the liquid is known as vapour pressure.
In a pure solvent, the vapour pressure is due to the solvent molecules. When a non-volatile solute is added to the solvent, the vapour pressure is only due to the solvent molecules. At a given temperature, the vapour pressure of solution is less than the vapour pressure of the pure solvent at the same temperature.
In the solution, the surface has both solvent and solute molecules, so the fraction of solvent molecules occupying the surface decreases. Due to this the number of solvent molecules escaping from the surface also decreases thus decreasing the vapour pressure of solution as compared to the pure solvent.
The decrease in vapour pressure of solvent depends on the quantity of non-pure solute present in the solution and not on its nature.
Raoult stated a law to give relation between vapour pressure of solvent and its mole fraction. The law stated that- For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
Consider a solution containing ${n_A}$ moles of solvent $A$ and ${n_B}$ moles of solute $B$ . Let ${p_A}$ be the vapour pressure of solvent in solution and $p_A^\circ $ be the vapour pressure of solvent in pure state. Then raoult law can be given as-
${p_A}\alpha {x_A}$
${p_A} = {x_A} \times p_A^\circ $
Where ${x_A}$ is the mole fraction of solvent in the solution.
Mole fraction of any component in the solution can be given as-
Mole fraction of an component \[ = {{number\,of\,moles\,of\,component}}{{Total\,number\,of\,moles\,of\,all\,components}}\]
In the given problem, let ${p_A}$ be the vapour pressure of water in aqueous solution and $p_A^\circ $ be the vapour pressure of water in its pure state. Then ${x_A}$ will be the mole fraction of water and ${x_B}$ will be the mole fraction of glucose that is added to water(here glucose is the solute).
First we need to calculate the mole fraction of water in the solution. It can be calculated as-
Mole fraction of water ${x_A} = {{{n_A}}}{{{n_A} + {n_B}}}$
Where ${n_A}$ are the number of moles of water present in the solution and ${n_B}$ are the number of moles of glucose present in the solution.
We know that $18g$ glucose is taken. So the number of moles of glucose are-
$number\,of\,moles({n_B}) = {{given\,mass}}{{Molecular\,mass\,of\,glu\cos e}} = {{18}}{{180}} = 0.1moles$
Similarly number of moles of water are-
$number\,of\,moles({n_A}) = {{given\,mass}}{{Molecular\,mass\,of\,water}} = {{178.2}}{{18}} = 9.9moles$
Now mole fraction of water present in the given solution is-
Mole fraction ${x_A} = {{{n_A}}}{{{n_A} + {n_B}}} = {{9.9}}{{9.9 + 0.1}} = 0.99$
Now it is given that the solution is at a temperature of ${100^\circ }C$ . this temperature is the boiling point of water. At this temperature the vapour pressure of pure water is equal to the atmospheric pressure $ = 760mmHg = 760torr$ .
So, $p_A^\circ $ $ = 760mmHg = 760torr$ .
Applying the Raoult’s law-
${p_A} = {x_A} \times p_A^\circ $
${p_A} = 0.99 \times 760 = 752.4torr$

So, the correct answer is Option D.

Note: The molecular formula of glucose is ${C_6}{H_{12}}{O_6}$ . The molecular mass can be calculated by the summation of atomic masses of all atoms.
The vapour pressure of solvent decreases when a non-volatile solute is added to it. This is called the relative lowering of vapour pressure. It depends only on the number of solute particles.