What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water (in $mm \ Hg$)? $(P_1^0 = 400 \ mm \ Hg)$

When a non-volatile,non-electrolyte solute is dissolved in a solvent,the vapor pressure of the solution decreases by $8\%$. The mole fraction of the solute is .............

When $25 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1} \ mm$. If the vapour pressure of water at $20^{\circ}C$ is $17.5 \ mm$, what is the molecular weight of the solute?

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 $\text{ atm}$)

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 $CCl_4 = 1.58 \ g/cm^3$)

What is the vapour pressure of a solution containing $0.1 \ mol$ of solute dissolved in $1.8 \times 10^{-2} \ kg$ of $H_2O$ (in $mm \ Hg$)? $(P_1^0 = 24 \ mm \ Hg)$