If a solution is prepared using the liquids mentioned in the previous question such that the mole fraction of $A$ is $0.8$,what will be the mole fraction of $A$ in the vapor phase?

The vapor pressure of an aqueous solution at $373 \, K$ is $0.925 \, atm$. What is the mole fraction of the solute?

Vapour pressure of solvent is $17.5 \ mm \ Hg$ while that of dilute solution is $17.45 \ mm \ Hg$; the mole fraction of solvent is

$A$ solution contains a $1:4$ molar ratio of pentane and hexane. At $20^{\circ}C$,the vapor pressure of pure pentane is $440 \ mm \ Hg$ and that of hexane is $120 \ mm \ Hg$. What is the mole fraction of pentane in the vapor phase?

$224 \, mL$ of $SO_{2(g)}$ at $273 \, K$ and $1 \, atm$ is passed through $100 \, mL$ of $0.1 \, M \, NaOH$ solution. The non-volatile solute produced is dissolved in $36 \, g$ of water. The lowering of vapour pressure of the solution (assuming the solution is dilute) $(P^0_{H_2O} = 24 \, mm \, Hg)$ is $x \times 10^{-2} \, mm \, Hg$. The value of $x$ is ...... . (Integer answer)

In a solution,the mole ratio of pentane $(A)$ and hexane $(B)$ is $1:4$. At $20^o C$,the vapor pressures of these pure hydrocarbons are $440 \ mm$ and $120 \ mm$ respectively. The mole fraction of pentane in the vapor phase is.....