Find the molar mass of a nonvolatile solute when $20 \ g$ of it is dissolved in $200 \ g$ of water at $300 \ K$. [Relative lowering of vapour pressure $= 0.02$]

At $298 \ K$,if the vapour pressure of pure liquids toluene,benzene,chloroform and dichloromethane are $60, 160, 200$ and $415 \ torr$ respectively,then which liquid has the highest boiling point?

At $298 \ K$,$0.714$ moles of liquid $A$ is dissolved in $5.555$ moles of liquid $B$. The vapour pressure of the resultant solution is $475 \ torr$. The vapour pressure of pure liquid $A$ at the same temperature is $280.7 \ torr$. What is the vapour pressure of pure liquid $B$ in $torr$?

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

Vapour pressure of a solution of $5 \ g$ of non-electrolyte in $100 \ g$ of water at a particular temperature is $2985 \ N/m^2$. The vapour pressure of pure water is $3000 \ N/m^2$. The molecular weight of the solute is

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