The plot of total vapour pressure as a function of mole fraction of the components of an ideal solution formed by mixing liquids $X$ and $Y$ is

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

In a solution,the mole fraction of solute is $0.2$ when the lowering in vapour pressure is $10 \text{ mm Hg}$. To achieve a lowering of vapour pressure of $20 \text{ mm Hg}$,the mole fraction of the solute in the solution should be:

Why is the vapour pressure of an aqueous solution of glucose lower than that of pure water?

The vapour pressure of water is $12.3 \, kPa$ at $300 \, K$. Calculate the vapour pressure of a $1 \, molal$ solution of a non-volatile solute in it.

An aqueous solution containing $28\%$ by mass of a liquid $A$ (molar mass $= 140 \ g/mol$) has a vapour pressure of $160 \ mm$ at $37 \ ^\circ C$. Find the vapour pressure of the pure liquid $A$ approximately .......... $mm$. (The vapour pressure of pure water at $37 \ ^\circ C$ is $150 \ mm$)