If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

The vapour pressure of pure $CCl_4$ (molar mass $= 154 \ g \ mol^{-1}$) and $SnCl_4$ (molar mass $= 170 \ g \ mol^{-1}$) at $25^{\circ} C$ are $115.0 \ torr$ and $238.0 \ torr$ respectively. Assuming ideal behaviour, calculate the total approximate vapour pressure in $torr$ of a solution containing $10 \ g$ of $CCl_4$ and $15 \ g$ of $SnCl_4$.

The vapour pressure of water is $12.3 \, kPa$ at $300 \, K$. Calculate the vapour pressure of a $1 \, molal$ solution of a non-volatile solute in it.

The vapour pressure of $30 \%$ $(w/v)$ aqueous solution of glucose is $...... \ mm \ Hg$ at $25^{\circ} \ C$. [Given : The density of $30 \%$ $(w/v)$ aqueous solution of glucose is $1.2 \ g \ cm^{-3}$ and vapour pressure of pure water is $24 \ mm \ Hg$.] (Molar mass of glucose is $180 \ g \ mol^{-1}$.)

The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?