Give the formula for the boiling point elevation constant $(K_b)$.
- A$K_b = \frac{R \times M_1 \times T_b^2}{1000 \times \Delta_{vap}H}$
- B$K_b = \frac{R \times M_1 \times T_b}{1000 \times \Delta_{vap}H}$
- C$K_b = \frac{R \times T_b^2}{1000 \times \Delta_{vap}H}$
- D$K_b = \frac{M_1 \times T_b^2}{1000 \times \Delta_{vap}H}$
Explore More
Similar Questions
The molal elevation constant of water is $0.513 \ ^oC \ kg \ mol^{-1}$. If $0.1 \ mol$ of sugar is dissolved in $200 \ g$ of water,at what temperature $(^oC)$ will the solution start to boil under $1 \ bar$ pressure?
Medium
View SolutionGive the unit for the molal elevation constant in boiling point.
Easy
View Solution$A$ solution containing $2.5 \times 10^{-3} \ kg$ of a solute dissolved in $75 \times 10^{-3} \ kg$ of water boils at $373.535 \ K$. The molar mass of the solute is $..... \ g \ mol^{-1}$. [nearest integer] (Given: $K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}$,boiling point of water $= 373.15 \ K$)
Elevation in boiling point was $0.52 \, ^\circ C$ when $6 \, g$ of a compound $X$ was dissolved in $100 \, g$ of water. The molecular weight of $X$ is: ($K_b$ for water is $0.52 \, K \, kg \, mol^{-1}$)
Medium
View SolutionThe latent heat of vaporisation of water is $9700 \, Cal/mole$ and if the boiling point is $100 \, ^oC$,the ebullioscopic constant of water is .......... $^oC \, kg/mol$.
MediumAIPMT 1989
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo