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:

When a non-volatile solute is added to a pure solvent,the vapor pressure decreases by $11.5 \ torr$. If the mole fraction of the solute is $0.2$,what will be the vapor pressure of the pure solvent in $torr$?

The vapor pressure of an aqueous solution at $373 \, K$ is $0.925 \, atm$. What is the mole fraction of the solute?

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

What is the relation between the vapour pressure of the solution,the vapour pressure of the pure solvent,and its mole fraction in the solution?

At $298 \ K$,the vapour pressure of a solution of $7.5 \ g$ of non-volatile solute in $90 \ g$ of water is $2.8 \ kPa$. If $18 \ g$ of water is added to this solution,the vapour pressure becomes $2.81 \ kPa$ at the same temperature. The molar mass of the solute in $g \ mol^{-1}$ is: