When $18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water,the vapour pressure of the aqueous solution at $100 \ ^\circ C$ is .......... $torr$.

"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

At $25\,^oC$,the vapour pressure of pure liquid $A$ (mol. wt. $= 40$) is $100\,torr$,while that of pure liquid $B$ is $40\,torr$ (mol. wt. $= 80$). The vapour pressure at $25\,^oC$ of a solution containing $20\,g$ of each $A$ and $B$ is ........... $torr$.

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

What is the vapour pressure of a solution containing $1.8 \ g$ of glucose in $16.2 \ g$ of water (in $mm \ Hg$)? ($P_1^0 = 24 \ mm \ Hg$ and molar mass of glucose $= 180 \ g \ mol^{-1}$)

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