What is molal elevation? Explain.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The elevation in boiling point is defined as the increase in the boiling point of a solution when $1 \ mol$ of a non-volatile solute is dissolved in $1 \ kg$ of solvent.
The boiling point of a solution is always higher than that of the pure solvent. This elevation depends on the number of solute particles rather than their nature.
Let $T_{b}^{0}$ be the boiling point of the pure solvent and $T_{b}$ be the boiling point of the solution. The increase in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{0}$.
For dilute solutions,the elevation of boiling point $(\Delta T_{b})$ is directly proportional to the molal concentration $(m)$ of the solute:
$\Delta T_{b} \propto m$
$\Delta T_{b} = K_{b} \times m$
Where $K_{b}$ is the Boiling Point Elevation Constant or Ebullioscopic Constant. Its unit is $K \ kg \ mol^{-1}$.
If $w_{2}$ grams of solute with molar mass $M_{2}$ are dissolved in $w_{1}$ grams of solvent,the molality $(m)$ is:
$m = \frac{w_{2} \times 1000}{M_{2} \times w_{1}}$
Substituting $m$ into the equation for $\Delta T_{b}$:
$\Delta T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$
Rearranging to find the molar mass of the solute:
$M_{2} = \frac{1000 \times w_{2} \times K_{b}}{\Delta T_{b} \times w_{1}}$

Explore More

Similar Questions

$180 \ g$ 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 $6 \ g$ of solute in $100 \ g$ of water boils at $100.52^{\circ} C$. The molal elevation constant of water is $0.52 \ K \ kg \ mol^{-1}$. What is the molar mass of the solute?

Calculate the number of moles of nonvolatile solute dissolved in $0.3 \ kg$ solvent if $\Delta T_{b}=0.3 \ K$ and $K_{b}$ for solvent is $1.8 \ K \ kg \ mol^{-1}$.

Calculate the molality of a solution containing a non-volatile solute if the boiling point elevation of the solution is $0.2 \ K$ $[K_{b} = 0.52 \ K \ kg \ mol^{-1}]$.

Calculate the molality of a solution of a non-volatile solute if the boiling point elevation is $1.75 \ K$ and the $K_b$ for the solvent is $3 \ K \ kg \ mol^{-1}$. (in $m$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo