The vapour pressure of pure liquid $A$ is $0.80 \ atm$. On mixing a non-volatile solute $B$ to $A$,its vapour pressure becomes $0.6 \ atm$. The mole fraction of $B$ in the solution is

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

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).

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 $(Y_{A})$?

State Raoult's law for a solution containing a non-volatile solute and a volatile solvent and explain it.

The vapor pressure of water at $23\,^{\circ}C$ is $19.8\,mm\,Hg$. What will be the vapor pressure (in $mm\,Hg$) of the resulting solution when $0.1\,mol$ of glucose is dissolved in $178.2\,g$ of water?