The vapour pressure of pure benzene and toluene are $160 \ torr$ and $60 \ torr$ respectively. The mole fraction of toluene in the vapour phase in contact with an equimolar solution of benzene and toluene is:

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water, what is the vapour pressure (in $mm$) of the resultant solution?

The vapour pressure of water is $12.3 \, kPa$ at $300 \, K$. Calculate the vapour pressure of a $1 \, molal$ solution of a non-volatile solute in 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?

$9 \ g$ anhydrous oxalic acid (mol. $Wt. = 90$) was dissolved in $9.9 \ moles$ of water. If vapour pressure of pure water is $P_1^o$,the vapour pressure of solution is (in $P_1^o$)

The vapor pressure of pure $A$ is $10 \ torr$. When $1 \ g$ of $B$ is dissolved in $20 \ g$ of $A$ at the same temperature,the vapor pressure of the solution becomes $9 \ torr$. If the molar mass of $A$ is $200 \ g/mol$,the molar mass of $B$ is ............ $amu$.