For two components $A$ and $B$,$P_A^0 : P_B^0 = 1:2$ and $X_A : X_B = 1:2$. The mole fraction of $A$ in the vapor phase is:

Vapour pressure of water at $293 \, K$ is $17.535 \, mm \, Hg$. Calculate the vapour pressure of water at $293 \, K$ when $25 \, g$ of glucose is dissolved in $450 \, g$ of water.

An aqueous solution of glucose was prepared by dissolving $18 \ g$ of glucose in $90 \ g$ of water. The relative lowering in vapour pressure is

The mass of non-volatile,non-electrolyte solute (molar mass $= 50 \ g \ mol^{-1}$) needed to be dissolved in $114 \ g$ octane to reduce its vapour pressure to $75\%$ is .............. $g$.

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

$A$ solution containing $30 \,g$ of non-volatile solute in $90 \,g$ of water has a vapour pressure of $2.8 \,kPa$ at $298 \,K$. Further,$18 \,g$ of water is added to the solution and the new vapour pressure becomes $2.9 \,kPa$ at $298 \,K$. Calculate:
$i$. Molar mass of the solute
$ii$. Vapour pressure of water at $298 \,K$