Relative Lowering of Vapour Pressure

When a certain amount of solid $\mathrm{A}$ is dissolved in $100 \mathrm{g}$ of water at $25 °\mathrm{C}$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76\mathrm{mmHg}$. The number of moles of solute $\mathrm{A}$ added is

$224 \mathrm{mL}$ of ${\mathrm{SO}}_{2\left(\mathrm{g}\right)}$ at $298 \mathrm{K}$ and $1 \mathrm{atm}$ is passed through $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ solution. The non-volatile solute produced is dissolved in $36 \mathrm{g}$ of water. The lowering of vapour pressure of solution (assuming the solution is dilute) ($\mathrm{P}{°}_{\left({\mathrm{H}}_2\mathrm{O}\right)}=24 \mathrm{mm}$ of $\mathrm{Hg}$) is $\mathrm{x}\times {10}^{-2} \mathrm{mm}$ of $\mathrm{Hg}$ the value of $\mathrm{x}$ is                (Integer answer)

A set of solutions is prepared using $180 \mathrm{g}$ of water as a solvent and $10 \mathrm{g}$ of different non-volatile solutes $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of $\mathrm{A}=100 \mathrm{g} {\mathrm{mol}}^{-1} ; \mathrm{B}=200 \mathrm{g} {\mathrm{mol}}^{-1} ; \mathrm{C}=10,000 \mathrm{g} {\mathrm{mol}}^{-1}$]

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:

At room temperature, a dilute solution of urea is prepared by dissolving $0.60\mathrm{ }\mathrm{g}$ of urea in $360\mathrm{ }\mathrm{g}$ of water. If the vapour pressure of pure water at this temperature is $35\mathrm{ }\mathrm{m}\mathrm{m}\mathrm{ }\mathrm{H}\mathrm{g}$ , lowering of vapour pressure will be:
(molar mass of urea $=60\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$ )

The mass of a non-volatile, non-electrolyte solute (molar mass $=\mathrm{ }50\mathrm{ }\mathrm{g}\mathrm{ }{\mathrm{mol}}^{-1}$) needed to be dissolved in $114 \mathrm{g}$ octane to reduce its vapour pressure by $75\%$, is:

$18 \mathrm{g}$ glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ is added to $178.2 \mathrm{g}$ water. The vapour pressure of water (in torr) for this aqueous solution is:

The vapour pressure of acetone at ${20}^{\mathrm{o}}\mathrm{C}$ is $185 \mathrm{torr}$. When $1.2 \mathrm{g}$ of a non-volatile substance was dissolved in $100 \mathrm{g}$ of acetone at ${20}^{\mathrm{o}}\mathrm{C}$ , its vapour pressure was $183 \mathrm{torr}$. The molar mass $\left(\mathrm{g}\mathrm{ }\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}\right)$ of the substance is: