$A$ 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 $150 \ g \ mol^{-1}$.
- A$1.95 \ K \ kg \ mol^{-1}$
- B$2.23 \ K \ kg \ mol^{-1}$
- C$1.52 \ K \ kg \ mol^{-1}$
- D$2.72 \ K \ kg \ mol^{-1}$
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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 $= 180 \ g \ mol^{-1}$) per $dm^3$ 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 $K$)
MediumKCET 2025
View SolutionWhat is the boiling point of a decimolal aqueous solution of glucose if the molal elevation constant for water is $0.52 \ ^{\circ}C \ kg \ mol^{-1}$ (in $^{\circ}C$)?
$58.4 \ g$ of $NaCl$ and $180 \ g$ of glucose were separately dissolved in $1000 \ mL$ of water. Identify the correct statement regarding the elevation of boiling point $(b.p.)$ of the resulting solutions.
DifficultWBJEE 2012
View SolutionThe value of $K_b$ is given by $mRT_b^2 / 1000 \, X$. In this relation,$X$ is :-
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