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 in $mm$ of $Hg$ of an aqueous solution obtained by adding $18 \ g$ of glucose $(C_6H_{12}O_6)$ to $180 \ g$ of water at $100^{\circ}C$ is:

The vapour pressure of water at $20\,^{\circ}C$ is $17.5\, mm\, Hg.$ If $18\, g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2\, g$ of water at $20\,^{\circ}C,$ the vapour pressure of the resulting solution will be $.........\, mm$ of $Hg$.

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 $180 \ g \ mol^{-1}$.)

Which one of the following solutions has the maximum vapour pressure at $27\,^{\circ}C$ temperature?

What is the mass in $g$ of a non-volatile solute with a molecular weight of $40$ that should be dissolved in $57 \ g$ of octane to reduce its vapour pressure to $80 \%$ of its original value?