$1 \ mol$ 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 .......... $mm \ Hg$.

At $298 \ K$,$0.714$ moles of liquid $A$ is dissolved in $5.555$ moles of liquid $B$. The vapour pressure of the resultant solution is $475 \ torr$. The vapour pressure of pure liquid $A$ at the same temperature is $280.7 \ torr$. What is the vapour pressure of pure liquid $B$ in $torr$?

Two beakers $A$ and $B$ containing pure water and an aqueous solution,respectively,are placed in a closed container. What will be observed gradually with the passage of time?

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

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

$A$ solution is obtained by dissolving $12 \ g$ of urea (mol.wt. $60$) in a litre of water. Another solution is obtained by dissolving $68.4 \ g$ of cane sugar (mol.wt. $342$) in a litre of water at the same temperature. The lowering of vapour pressure in the first solution is