At a temperature of $300 \ K$,if a liquid is transferred from a small test tube to a large beaker,what will be the change in its vapor pressure?

$A$ 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 $5 \times 10^{-3}$?

$A$ solution of a non-volatile solute is obtained by dissolving $2 \ g$ of solute in $50 \ g$ of benzene. Calculate the vapour pressure of the solution if the vapour pressure of pure benzene is $640 \ mmHg$ at $25^{\circ} C$. [Molar mass of benzene $= 78 \ g \ mol^{-1}$,Molar mass of solute $= 64 \ g \ mol^{-1}$] (in $mm \ Hg$)

Calculate the relative lowering of vapour pressure of a solution containing $46 \ g$ of non-volatile solute in $162 \ g$ of water at $20^{\circ} C$. [Molar mass of non-volatile solute $= 46 \ g \ mol^{-1}$]

Vapour pressure of a pure solvent is $550 \ mm$ of $Hg$. By addition of a non-volatile solute,it decreases to $510 \ mm$ of $Hg$. Calculate the mole fraction of the solute in the solution.

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.