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

Lowering of vapour pressure due to a solute in $1 \, molal$ aqueous solution at $100 \, ^\circ C$ is ........ $torr$.

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 $mm \ Hg$)? $(P_1^0 = 24 \ mm \ Hg)$

$224 \, mL$ of $SO_{2(g)}$ at $273 \, K$ and $1 \, atm$ is passed through $100 \, mL$ of $0.1 \, M \, NaOH$ solution. The non-volatile solute produced is dissolved in $36 \, g$ of water. The lowering of vapour pressure of the solution (assuming the solution is dilute) $(P^0_{H_2O} = 24 \, mm \, Hg)$ is $x \times 10^{-2} \, mm \, Hg$. The value of $x$ is ...... . (Integer answer)

Calculate the relative lowering of vapour pressure if the vapour pressure of the pure solvent and the vapour pressure of the solution at $25^{\circ} C$ are $32 \ mm \ Hg$ and $30 \ mm \ Hg$ respectively.

At $300 \ K$, $0.06 \ kg$ of an organic solute is dissolved in $1 \ kg$ water. The vapour pressure of the solution at $300 \ K$ is $3.768 \ kPa$. If the vapour pressure of water at that temperature is $3.78 \ kPa$, what is the molar mass of the organic solute (in $g \ mol^{-1}$)? (Assume the solution is dilute)