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 .............. $g$.

$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.

$18 \ g$ of glucose is dissolved in $90 \ g$ of water. The relative lowering of vapour pressure of the solution is equal to

The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm$ and $700 \ mm$ of $Hg$ at $350 \ K$ respectively. If the total vapour pressure of the mixture is $600 \ mm$ of $Hg$,the composition of the mixture in the solution is

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 $\text{ atm}$)

Vapour pressure of a pure solvent at $90\,^{\circ}C$ is $1020\, torr$. $A$ solution of $0.1\, m$ concentration has a vapour pressure of $1000\, torr$. What is the molecular weight of the solvent?