For a dilute solution containing a nonvolatile solute, the molar mass of solute evaluated from the elevation of boiling point is given by the expression:

M_{2} = \frac{Δ T_{b}}{K_{b}} \frac{m_{1}}{m_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M_{2} = \frac{Δ T_{b}}{K_{b}} \frac{m_{2}}{m_{1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M_{2} = \frac{K_{b}}{Δ T_{b}} \frac{m_{2}}{m_{1}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

M_{2} = \frac{K_{b}}{Δ T_{b}} \frac{m_{1}}{m_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $M_{2} = \frac{K_{b}}{Δ T_{b}} \frac{m_{2}}{m_{1}}$
Elevation of Boiling-point describes that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. This happens whenever a non-volatile solute, such as a salt, is added to a pure solvent, such as water.

Elevation of boiling point ( $Δ T_{b}$ ) is the difference between the boiling points of the solution and the pure solvent:

$Δ T_{b} = T_{b} - {T_{b}}^{0}$

Elevation of boiling point can be calculated as $Δ T_{b} = m . K_{b}$

where m is the concentration of the solute expressed in molality and $K_{b}$ is the molal boiling point elevation constant of the solvent.

$m = \frac{\frac{m_{2}}{M_{2}}}{m_{1}}$

$m_{2}$ = mass of solute
$m_{1}$ = mass of solvent
$M_{2}$ = molecular wt. of solute.

From $Δ T_{b} = m . K_{b}$

We get,

$M_{2} = \frac{K_{b}}{Δ T_{b}} . \frac{m_{2}}{m_{1}}$

Suggest Corrections

Similar questions

Q. Match the column I with column II and mark the appropriate choice

Column I	Column II
(A) $K_{b}$	(i) $\frac{K_{b} \times W_{2} \times 1000}{{Δ T_{b} \times W}_{1}}$
(B) $M_{2}$	(ii) $\frac{W_{2} \times 1000}{{M_{2} \times W}_{1}}$
(C) $π$	(iii) $\frac{R T_{b}^{2}}{1000 \times L_{V}}$
(D) $m$	(iv) $\frac{Δ T_{b} \times d R T}{1000 \times K_{b}}$

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is: