Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

Vapour pressure of a solution of $5 \ g$ of non-electrolyte in $100 \ g$ of water at a particular temperature is $2985 \ N/m^2$. The vapour pressure of pure water is $3000 \ N/m^2$. The molecular weight of the solute is

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 $(Y_{A})$?

$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

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 ............... $g/mol$.

$A$ mixture of volatile components $A$ and $B$ has a total vapour pressure (in $torr$) $P = 254 - 119 X_A$,where $X_A$ is the mole fraction of $A$ in the mixture. What are the values of $P_A^o$ and $P_B^o$ (in $torr$)?