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 $20^\circ C$,what will be the vapour pressure of the resulting solution (in $mm \ Hg$)?

"Relative lowering in vapour pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solute". The above statement is:

In a liquid mixture of $A$ and $B$,what is the mole fraction of $A$ in the vapor phase $(X_A = 0.4)$? Given: $P_A^o = 100 \, mm \, Hg$ and $P_B^o = 200 \, mm \, Hg$.

One mole of a non-volatile solute is dissolved in two moles of water. What is the vapor pressure of this solution relative to that of pure water?

When a solute is added to a solvent having a vapor pressure of $0.80 \ atm$,the vapor pressure decreases to $0.60 \ atm$. The mole fraction of the solute is .........

At $298 \ K$,the vapour pressure of pure water is $25 \ torr$. The vapour pressure of water,when $12 \ g$ of urea (molar mass,$60 \ g \ mol^{-1}$) and $36 \ g$ of glucose (molar mass,$180 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of water at the same temperature (in $torr$) is