Calculate the molar mass of a non-volatile solute when $5 \ g$ of it is dissolved in $50 \ g$ of solvent,which boils at $119.6^{\circ} C$. $[K_{b} = 3.2 \ K \ kg \ mol^{-1}$,boiling point of pure solvent $= 118^{\circ} C]$.

When a substance is dissolved in a solvent,the vapour pressure of the solvent is decreased. This results in:

When $1.8 \ g$ of glucose is dissolved in $100 \ g$ of solvent,the boiling point of the solution increases by $0.1 \ ^\circ C$. The molal elevation constant of the solvent is ............ $\frac{K}{m}$.

The molal elevation boiling point constant for water is $0.513 \ ^{\circ}C \ kg \ mol^{-1}$. Calculate the boiling point of the solution if $0.1 \ mol$ of sugar is dissolved in $200 \ g$ of water. (in $^{\circ}C$)

$18 \, g$ of glucose $(C_{6}H_{12}O_{6})$ dissolved in $1 \, kg$ of water is heated to boiling. The boiling point (in $K$) measured at $1 \, atm$ pressure is closest to: [Ebullioscopic constant,$K_{b}$ for water is $0.52 \, K \, kg \, mol^{-1}$. Consider the boiling point of pure water to be $373.15 \, K$]

The plot given below shows $P-T$ curves (where $P$ is the pressure and $T$ is the temperature) for two solvents $X$ and $Y$ and isomolal solutions of $NaCl$ in these solvents. $NaCl$ completely dissociates in both the solvents.
On addition of equal number of moles of a non-volatile solute $S$ in equal amount (in $kg$) of these solvents,the elevation of boiling point of solvent $X$ is three times that of solvent $Y$. Solute $S$ is known to undergo dimerization in these solvents. If the degree of dimerization is $0.7$ in solvent $Y$,the degree of dimerization in solvent $X$ is. . . . . . .