$60 \ g$ of Urea (Mol. wt $60$) was dissolved in $9.9 \ moles$ of water. If the vapour pressure of pure water is $P_o$,the vapour pressure of the solution is: (in $P_o$)

In a liquid mixture of $A$ and $B$,what is the mole fraction of $A$ in the vapor phase $(X_A = 0.4)$? Given: $P_A^o = 100 \, mm \, Hg$ and $P_B^o = 200 \, mm \, Hg$.

Two liquids $A$ and $B$ have $P_A^o$ and $P_B^o$ in the ratio of $1 : 3$ and the ratio of number of moles of $A$ and $B$ in the liquid phase is $1 : 3$. The mole fraction of $A$ in the vapour phase in equilibrium with the solution is equal to:

When a certain amount of solid $A$ is dissolved in $100 \ g$ of water at $25^{\circ} \ C$ to make a dilute solution,the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76 \ mm \ Hg$. The number of moles of solute $A$ added is $...........$ (Nearest Integer).

$A$ solution is prepared by mixing $8.5 \ g$ of $CH_2Cl_2$ and $11.95 \ g$ of $CHCl_3$. If the vapour pressures of pure $CH_2Cl_2$ and $CHCl_3$ at $298 \ K$ are $415 \ mm \ Hg$ and $200 \ mm \ Hg$ respectively,the mole fraction of $CHCl_3$ in the vapour phase is: (Molar mass of $Cl = 35.5 \ g \ mol^{-1}$)

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)