What is the vapour pressure of a solution containing $0.1 \text{ mol}$ of non-volatile solute dissolved in $16.2 \text{ g}$ water (in $\text{ mmHg}$)? ($P_1^0 = 24 \text{ mmHg}$, molar mass of water $18 \text{ g mol}^{-1}$)

$A$ mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 \, mm$ at $300 \, K$. The vapour pressure of propyl alcohol is $200 \, mm$. If the mole fraction of ethyl alcohol is $0.6$,its vapour pressure (in $mm$) at the same temperature will be:

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

The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

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.

According to Raoult's law for a non-volatile solute,which of the following is correct?