Vapour pressure of a solution of $5 \ g$ of non-electrolyte in $100 \ g$ of water at a particular temperature is $2985 \ N/m^2$. The vapour pressure of pure water is $3000 \ N/m^2$. The molecular weight of the solute is

$A$ 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 $298 \,K$

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

At $T \ K$,the vapour pressure of an $x$ molal aqueous solution containing a non-volatile solute is $12.078 \ kPa$. The vapour pressure of pure water at $T \ K$ is $12.3 \ kPa$. What is the value of $x$?

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 ............ $amu$.

$A$ solution is prepared by mixing $8.5 \ g$ of $CH_2Cl_2$ and $11.95 \ g$ of $CHCl_3$. If the vapour pressures of pure $CH_2Cl_2$ and $CHCl_3$ at $298 \ K$ are $415 \ mm \ Hg$ and $200 \ mm \ Hg$ respectively,the mole fraction of $CHCl_3$ in the vapour phase is: (Molar mass of $Cl = 35.5 \ g \ mol^{-1}$)