$18 \ g$ of glucose,$C_6H_{12}O_6$,is dissolved in $1 \ kg$ of water in a saucepan. At what temperature will water boil at $1.013 \ bar$ (in $K$)? $K_b$ for water is $0.52 \ K \ kg \ mol^{-1}$.

Calculate the boiling point of an aqueous solution containing $18 \text{ g}$ of glucose in $100 \text{ g}$ of water, given that the molal elevation constant $(K_b)$ of water is $0.5 \text{ K kg mol}^{-1}$. [Molar mass of glucose $= 180 \text{ g mol}^{-1}$ and boiling point of pure water $= 100^{\circ}\text{C}$] (in $^{\circ}\text{C}$)

Which of the following statements is correct for the boiling point of a liquid?

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:

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

The elevation in boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of solvent is $0.1^{\circ}C$. The value of $K_b$ for the solvent is ........ $\frac{K}{m}$.