If the boiling point of a solution is $T_1$ and the boiling point of the pure solvent is $T_2$,then the elevation in boiling point is given by:

$A$ solution of non-volatile solute has a boiling point elevation of $1.75 \ K$. Calculate the molality of the solution $[K_{b} = 3.5 \ K \ kg \ mol^{-1}]$.

$180 \ g$ 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 $K$)

Which of the following aqueous solutions will have a boiling point of $102.2^{\circ} C$? The molal elevation constant for water is $K_b = 0.512 \ K \ kg \ mol^{-1}$. (Note: The provided constant $2.2 \ K \ kg \ mol^{-1}$ in the prompt is incorrect for water; using standard $K_b = 0.512 \ K \ kg \ mol^{-1}$ for calculation).

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 ............ $\frac{K}{m}$.

When $10 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of benzene,it raises the boiling point by $1 \ ^{\circ}C$. The molecular mass of the solute is ....... $g \ mol^{-1}$. ($K_b$ for benzene $= 2.53 \ K \ kg \ mol^{-1}$)