$34.2 \ g$ of cane sugar is dissolved in $180 \ g$ of water. The relative lowering of vapour pressure will be

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

The vapour pressures of $A$ and $B$ at $25^{\circ} C$ are $90 \ mm \ Hg$ and $15 \ mm \ Hg$ respectively. If $A$ and $B$ are mixed such that the mole fraction of $A$ in the mixture is $0.6$,then the mole fraction of $B$ in the vapour phase is $x \times 10^{-1}$. The value of $x$ is $.....$ (Nearest integer)

Vapour pressure of $CCl_{4}$ at $25\,^{\circ} C$ is $143\, mm\, Hg$. $0.5\, g$ of a non-volatile solute (mol. wt. $65$) is dissolved in $100\, mL$ of $CCl_{4}$. Find the vapour pressure of the solution. (Density of $CCl_{4} = 1.58\, g / cm^{3}$)

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

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