You visited us 1 times! Enjoying our articles? Unlock Full Access!

Elevation in Boiling Point

Boiling point...

Question

Boiling point of a solvent is $80.2$ $^{o} C$ . When $0.419$ g of the solute of molar mass $252.4$ g $m o l^{- 1}$ was dissolved in $75$ g of the solvent, the boiling point of the solution was found to be ${80.256}^{o}$ C. Find the molal elevation constant.

1.25 K k g m o l^{- 1}

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

2.53 K k g m o l^{- 1}

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

2.87 K k g m o l e^{- 1}

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

None of these

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

Open in App

Solution

The correct option is B $2.53 K k g m o l^{- 1}$
Boiling point is the temperature when the vapour pressure of liquid equals the atmospheric pressure.
$Δ T_{b} =$ Boiling point of solution - Boiling point of solvent -------(1)
$Δ T_{b} = K_{b} \times m$
where,
$K_{b}$ is Ebullioscopic constant/ molal elevation constant
$T_{b (s o l)} - T_{b}^{o} = K_{b} \times m$
$80.256 - 80.2 = K_{b} \times \frac{m a s s o f s o l u t e \times 1000}{M o l a r m a s s o f s o l u t e \times M a s s o f s o l v e n t (i n g m)}$
$0.056 = K_{b} \times \frac{0.419 \times 1000}{252.4 \times 75}$

$K_{b} = 2.53 K k g m o l^{- 1}$

Suggest Corrections

Similar questions

Q. If 0.15 g of a solute, dissolved in 15 g of solvent, increases the boiling point by

{0.216}^{o} C

over that of the pure solvent, the molecular mass of the substance is (Molal elevation constant for the solvent is 2.16^OC$):

Q. If

1 g

of solute (molar mass

= 50 g

{m o l}^{- 1}

) is dissolved in

50 g

of solvent and the elevation in boiling point is

1 K

. The molar boiling constant of the solvent is?

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. $(K_{b} f o r b e n z e n e i s 2.53 K k g m o l^{- 1})$

Q. a) The boiling point of benzene is

353.23

K. When

1.80

g of a non-volatile non-ionisation solute was dissolved in

90

g of benzene, the boiling point raised to

354.11

Calculate the molar mass of the solute. [

K_{b}

for benzene

= 2.53

K Kg mol

^{- 1}

b) Define:

i. The molality of a solution.
ii. Isotonic solutions.

Q. If

30 g

of a solute of molecular weight

154 g m o l^{- 1}

is dissolved in

250 g

of benzene, what will be the boiling point of the resulting solution under atmospheric pressure?
The molal boiling-point elevation constant for benzene is

2.61 K k g m o l^{- 1}

and the boiling point of pure benzene is

{80.1}^{o} C .

Join BYJU'S Learning Program

Boiling point of a solvent is 80.2 oC. When 0.419g of the solute of molar mass 252.4g mol−1 was dissolved in 75g of the solvent, the boiling point of the solution was found to be 80.256oC. Find the molal elevation constant.

Boiling point of a solvent is $80.2$ $^{o} C$ . When $0.419$ g of the solute of molar mass $252.4$ g $m o l^{- 1}$ was dissolved in $75$ g of the solvent, the boiling point of the solution was found to be ${80.256}^{o}$ C. Find the molal elevation constant.