The vapour pressure of pure water is $23 \text{ mmHg}$. The vapour pressure of an aqueous solution, which contains $10$ mass per cent of solute '$A$' having molecular weight $50$ is (in $\text{ atm}$)

$34.2 \ g$ of cane sugar is dissolved in $180 \ g$ of water. The relative lowering of vapour pressure will be

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

Calculate the vapour pressure of the solution if the relative lowering of vapour pressure and the vapour pressure of the pure solvent are $0.018$ and $18 \ mm \ Hg$ respectively at $300 \ K$. (in $mm \ Hg$)

$A$ solution was prepared by dissolving $0.1 \ mol$ of a nonvolatile solute in $0.9 \ mol$ of water. What is the relative lowering of vapour pressure of the solution?

The variation of vapour pressure of different liquids with temperature is shown in the figure.
$(i)$ Calculate graphically the boiling points of liquids $A$ and $B$.
(ii) If we take liquid $C$ in a closed vessel and heat it continuously,at what temperature will it boil?
(iii) At high altitude,atmospheric pressure is low (say $60 \ mm \ Hg$). At what temperature does liquid $D$ boil?
(iv) $A$ pressure cooker is used for cooking food at hill stations. Explain in terms of vapour pressure why this is so.