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

When $25 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1} \ mm$. If the vapour pressure of water at $20^{\circ}C$ is $17.5 \ mm$, what is the molecular weight of the solute?

The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

At $T \ K$, $0.1 \ mol$ of a non-volatile solute was dissolved in $0.9 \ mol$ of a volatile solvent. The vapour pressure of pure solvent is $0.9 \ bar$. What is the vapour pressure (in $bar$) of the solution?

The vapour pressure of pure benzene at a certain temperature is $0.850 \ bar$. $A$ non-volatile, non-electrolyte solid weighing $0.5 \ g$ is added to $39.0 \ g$ of benzene (molar mass $78 \ g/mol$). The vapour pressure of the solution is then $0.845 \ bar$. What is the molar mass of the solid substance?

$A$ solution was prepared by dissolving $0.1 \ mol$ of a nonvolatile solute in $0.9 \ mol$ of water. What is the relative lowering of vapour pressure of the solution?