If $K_{b}$ denotes the molal elevation constant of water,then the boiling point of an aqueous solution containing $36 \ g$ of glucose (molar mass $= 180 \ g/mol$) per $1 \ kg$ of water is:

$180 \ g$ 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 $K$)

$A$ solution of a nonvolatile solute is obtained by dissolving $1.5 \ g$ in $30 \ g$ of solvent. The boiling point elevation is $0.65 \ K$. Calculate the molal elevation constant if the molar mass of the solute is $150 \ g \ mol^{-1}$.

When common salt is dissolved in water,what happens to the boiling point of the solution?

$A$ 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. $(K_b = 2.5 \ K \ kg \ mol^{-1})$

The boiling point of a solution of $0.11 \ g$ of a substance in $15 \ g$ of ether was found to be $0.1 \ ^\circ C$ higher than that of the pure ether. The molecular weight of the substance will be $(K_b = 2.16 \ K \ kg \ mol^{-1})$.