In $108 \, g$ of water,$18 \, g$ of a non-volatile compound is dissolved. At $100^{\circ} C$,the vapour pressure of the solution is $750 \, mm \, Hg$. Assuming that the compound does not undergo association or dissociation,the molar mass of the compound in $g \, mol^{-1}$ is

Consider a binary solution of two volatile liquid components $1$ and $2$. $x_1$ and $y_1$ are the mole fractions of component $1$ in liquid and vapour phase,respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as :

Calculate the vapour pressure of a solution containing a mixture of $2 \ mol$ of volatile liquid $A$ and $3 \ mol$ of volatile liquid $B$ at room temperature. $(P_{A}^{\circ} = 420 \ mm \ Hg, P_{B}^{\circ} = 610 \ mm \ Hg)$ (in $mm \ Hg$)

At $300 \ K$, $0.06 \ kg$ of an organic solute is dissolved in $1 \ kg$ water. The vapour pressure of the solution at $300 \ K$ is $3.768 \ kPa$. If the vapour pressure of water at that temperature is $3.78 \ kPa$, what is the molar mass of the organic solute (in $g \ mol^{-1}$)? (Assume the solution is dilute)

The vapour pressure of water at $20^\circ C$ is $17.5 \ mm \ Hg$. If $18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water at $20^\circ C$,what will be the vapour pressure of the resulting solution (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).