The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

Calculate the relative lowering of vapour pressure of a solution containing $3 \ g$ urea in $50 \ g$ water. [ Molar mass of urea $= 60 \ g \ mol^{-1} ]$

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 $\text{ atm}$)

Dry air was drawn through bulbs containing pure water,then through bulbs containing a solution of $600 \ g$ of a non-electrolyte in $360 \ g$ of water at the same temperature,and finally through a tube in which dried $CaCl_2$ was placed. The solution bulb gained $1.5 \ g$ and the dried $CaCl_2$ gained $2 \ g$. The molecular mass of the solute is:

The vapour pressure in $mm$ of $Hg$ of an aqueous solution obtained by adding $18 \ g$ of glucose $(C_6H_{12}O_6)$ to $180 \ g$ of water at $100^{\circ}C$ is:

At $T \ K$,the vapour pressure of water is $x \ kPa$. What is the vapour pressure (in $kPa$) of a $1 \ m$ (molal) aqueous solution containing a non-volatile solute (in $x$)?