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

$180 \ g$ 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 $K$)

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

Boiling point of a $2\%$ aqueous solution of a non-volatile solute $A$ is equal to the boiling point of an $8\%$ aqueous solution of a non-volatile solute $B$. The relation between molecular weights of $A$ and $B$ is:

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

$A$ solution of a nonvolatile solute is obtained by dissolving $3.5 \ g$ of solute in $100 \ g$ of solvent. The boiling point elevation is $0.35 \ K$. Calculate the molar mass of the solute. $(K_b = 2.5 \ K \ kg \ mol^{-1})$