The rise in the boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of a solvent is $0.1 \ ^\circ C$. The molal elevation constant of the liquid is .......... $K/m$.

$180 \ g$ of glucose,$C_6H_{12}O_6$,is dissolved in $1 \ kg$ of water in a vessel. The temperature at which water boils at $1.013 \ bar$ is $ . . . . . . $ (given,$K_b$ for water is $0.52 \ K \ kg \ mol^{-1}$. Boiling point for pure water is $373.15 \ K$) (in $K$)

If molality of a solution is $0.05 \ m$ and elevation in boiling point is $0.16 \ K$,then what is the molal elevation constant of the solvent?

The boiling point of toluene is $110.7\,^oC$ and its molal elevation constant is $3.32\, K\, kg\, mol^{-1}$. The enthalpy of vaporization of liquid toluene is ............ $kJ\, mol^{-1}$.

If $10 \ g$ of a solute (molar mass $= 100 \ g/mol$) is dissolved in $100 \ g$ of water,then the molal elevation constant $K_b$ of the solvent will be equal to:

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 $= 118^{\circ} C]$.