The vapour pressure of a dilute aqueous solution of glucose is $750 \ mm$ of mercury at $373 \ K$. The mole fraction of solvent is

When $5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the vapor pressure decreases from $3000 \ N \ m^{-2}$ to $2985 \ N \ m^{-2}$. The molar mass of the solute is ............... $g/mol$.

What is the vapour pressure of a solution containing $0.1 \ mol$ of non-volatile solute dissolved in $16.2 \ g$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^o : P_B^o = 1 : 3$ at a certain temperature. If the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4 : 3$,then the mole fraction of $B$ in the solution at the same temperature is

Assertion $:-$ Vapour pressure of a liquid increases on increasing surface area.
Reason $:-$ Rate of vaporisation of a liquid is directly proportional to surface area.

For two volatile liquids $A$ and $B$,if the vapour pressure ratio of $P_{A}^{0} : P_{B}^{0} = 1 : 2$ and the mole fraction ratio in the liquid phase is $X_{A} : X_{B} = 1 : 2$,then find the mole fraction of component $A$ in the vapour state $(Y_{A})$?