$A$ solution is prepared by dissolving $10 \ g$ of a non-volatile solute (molar mass,'$M$ $g \ mol^{-1}$') in $360 \ g$ of water. What is the molar mass in $g \ mol^{-1}$ of solute if the relative lowering of vapour pressure of solution is $5 \times 10^{-3}$?

Calculate the relative lowering of vapour pressure of a solution containing $0.56 \ g$ of a non-volatile solute in $100 \ g$ of water [molar mass of solute $= 60 \ g \ mol^{-1}$].

Assertion $(A)$: The vapour pressure of $0.1 \ M$ sugar solution is less than that of $0.1 \ M$ $KCl$ solution.
Reason $(R)$: Lowering of vapour pressure is directly proportional to the number of particles of non-volatile solute present in the solution.
The correct answer is

Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \ mmHg$ and $590 \ mmHg$ respectively at the same temperature.

$A$ solution containing $30 \,g$ of non-volatile solute in $90 \,g$ of water has a vapour pressure of $2.8 \,kPa$ at $298 \,K$. Further,$18 \,g$ of water is added to the solution and the new vapour pressure becomes $2.9 \,kPa$ at $298 \,K$. Calculate:
$i$. Molar mass of the solute
$ii$. Vapour pressure of water at $298 \,K$

At $25^\circ C$,the vapor pressure of $CCl_4$ is $143 \ mm \ Hg$. If $0.5 \ g$ of a non-volatile solute (molar mass $= 65 \ g/mol$) is dissolved in $100 \ mL$ of $CCl_4$,the vapor pressure of the resulting solution is $...... \ mm \ Hg$. (Density of $CCl_4 = 1.58 \ g/cm^3$)