Question

18 g of glucose (\[{C_6}{H_{12}}{O_6}\]) is added to 178.2 g of water. The vapour pressure of water (in torr) for this aqueous solution is:
A: 76.0
B: 752.4
C: 759.0
D: 7.6

Answer 1

Hint: Vapour pressure is actually a measure of the tendency of a substance to change into a gaseous or a vapour state, and to be noted that it increases with temperature.

Complete Step by step answer:
$Number{\text{ }}of{\text{ }}moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}mass(gmo{l^{ - 1}})}}$
Using the above formula we will calculate the number of moles of water and glucose.
Mass of water = 178.2 g (Given)
Molar mass of water \[\left( {{H_2}O} \right) = \left( {2 \times 1} \right) + \left( {1 \times 16} \right){\text{ = }}18{\text{ }}g\]
$\therefore Number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}{H_2}O = \dfrac{{178.2}}{{18}} = 9.9$
Mass of glucose = 18 g (Given)
Molar mass of glucose \[({C_6}{H_{12}}{O_6}) = (6 \times 12) + \left( {12 \times 1} \right) + \left( {6 \times 16} \right) = 180{\text{ }}g\]
$\therefore Number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}{{\text{C}}_6}{H_{12}}{O_6} = \dfrac{{18}}{{180}} = 0.1$
Hence, total number of moles = $9.9 + 0.1 = 10$moles
Now, it should be noted that mole fraction of glucose in the solution is equal to the change in pressure with respect to the total initial pressure or in simpler terms, we can say that the mole fraction can also be expressed as the ratio of the partial pressure to the total pressure of the solution.
(And mole fraction refers to the ratio of the number of moles linked to one component in a solution or mixture to the total number of moles)
Mole fraction of glucose XA can be calculated as follows:
${X_A} = \dfrac{{0.1}}{{10}} = 0.01$
$
  \therefore \dfrac{{\Delta p}}{{\Delta {p_o}}} = 0.01 \\
  \Delta p = 0.01 \times \Delta {p_o} = 0.01 \times 760 = 7.6\;torr \\
$
Thus, desired vapour pressure =$760 - 7.6 = 752.4\;torr$

Therefore, the correct answer is Option B.

Note: Raolt’s law can also be used to solve this question. According to Raoult's law, the vapour pressure of a component at a given temperature equals the mole fraction of that component times the vapour pressure of that component in a pure state.
Mole fraction of water XB$ = 1 - 0.01 = 0.99$
According to Raolt’s law:
${P_B} = {P_B}^0{X_B} = 760 \times 0.99 = 752.4$