Question

An aqueous solution of glucose is made by dissolving 10 g of glucose in 90g water at 300K. If the VP of pure water at 303 K be 32.6 mm Hg. What would be the VP of the solution?

Answer 1

Hint: Vapour pressure is the measure of the tendency of a material which changes the gaseous or the vapor state, and the vapor pressure increases with the increase in the temperature.

Complete step by step solution:
To find the vapour pressure of the solution you require the following information:
The molar mass of glucose = 180 g/mol
The molar mass of water = 18 g/mol
The number of moles of sugar i.e. glucose can be calculated by dividing the given mass to the molar mass. Given the question:
Given mass of glucose = 10 g
Number of moles of glucose = $\dfrac{given\text{ }mass}{molar\text{ mass}}=\dfrac{10}{180}=0.05mole$
Given mass of water = 90 g
Number of moles of water =$\dfrac{given\text{ }mass}{molar\text{ mass}}=\dfrac{90}{18}=5mole$
Now in the next step we will calculate the mole fraction of the resulting solution i.e. the mole fraction is calculated by the ratio of number of moles of water present to the total number of moles present in the solution i.e. the number of moles of water and the number of moles of glucose.
Mole fraction of solution =$\dfrac{5}{(5+0.5)}=\dfrac{5}{5.05}=0.99$
Not we will determine the vapour pressure of the resultant solution:
The vapour pressure of the solutions will be equal to the mole fraction of the solvent
VP of solution = $(0.99)(32.8)$ mm of Hg
= 32.74 mm of Hg

Note: The vapour pressure of a liquid varies with the change in temperature, different liquids have different values of vapour pressure at different temperatures. Another factor which affects the vapour pressure of a liquid is surface area and intermolecular forces.