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

$1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ is mixed with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$. The vapor pressure of the resulting ideal solution is .......... $mm \ Hg$.

Assertion $(A)$: The vapour pressure of $0.1 \ M$ sugar solution is less than that of $0.1 \ M$ $KCl$ solution.
Reason $(R)$: Lowering of vapour pressure is directly proportional to the number of particles of non-volatile solute present in the solution.
The correct answer is

An aqueous solution is $1.00 \ molal$ in $KI$. Which change will cause the vapour pressure of the solution to increase?

$A$ solution is prepared by dissolving $68 \text{ g}$ of sucrose $(C_{12}H_{22}O_{11})$ in $1 \text{ kg}$ of water. Calculate the vapour pressure of the solution at $298 \text{ K}$. Given: Vapour pressure of pure water at $298 \text{ K} = 18 \text{ mm Hg}$ and mole fraction of solvent $(x_1)$ $= 0.9964$. (in $\text{ mm Hg}$)

"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: