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 $= 180 \ g \ mol^{-1}$)

The vapor pressures of two pure liquids $(A)$ and $(B)$ are $100 \ Torr$ and $80 \ Torr$ respectively. What is the total vapor pressure of the solution formed by mixing $2 \ moles$ of $(A)$ and $3 \ moles$ of $(B)$ (in $Torr$)?

$3 \ g$ of urea is dissolved in $45 \ g$ of $H_2O$. The relative lowering in vapour pressure is

Two liquids $A$ and $B$ form an ideal solution. At $320 \ K$, the vapour pressure of the solution, containing $3 \ mol$ of $A$ and $1 \ mol$ of $B$ is $500 \ mm \ Hg$. At the same temperature, if $1 \ mol$ of $A$ is further added to this solution, the vapour pressure of the solution increases by $20 \ mm \ Hg$. The vapour pressure (in $mm \ Hg$) of $B$ in the pure state is . . . . . . (Nearest integer).

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 120 - 80 \, X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and $B$ at the same temperature are respectively:

The relative lowering of vapor pressure is equal to the mole fraction of the non-volatile solute. This statement was given by whom?