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Elevation in Boiling Point

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Question

A solution of $2.5$ g of a non-volatile solid in $100$ g benzene is boiled at ${0.42}^{o} C$ higher than the boiling point of pure benzene. Calculate the molecular mass of the substance. Molal elevation constant of benzene is $2.67 K . k g . m o l^{- 1}$

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Solution

$M = \frac{1000 K_{b} \times w}{W \times △ T}$

Given,
$K_{b} = 2.67 K . k g . m o l^{- 1},$
$w = 2.5 g,$
$W = 100 g,$
$△ T = 0.42 K$

Now,

$M = \frac{1000 \times 2.67 \times 2.5}{100 \times 0.42} = 158.9$

The molecular weight of substance is $158.9 g / m o l$

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Similar questions

Q. Y g of non-volatile organic substance of molar mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is

K_{b}

. Elevation in its boiling point is given by :

y

gm of a non-volatile organic substance of molecular mass

M

is dissolved in 250 gm benzene. The molal elevation constant of benzene is

K_{b}

. The elevation in its boiling point is given by:

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 .

Q. Boiling point of benzene is

353.23

1.8

g of non-volatile solute is dissolved in

90

g of benzene. Then boiling point raised to

354.11

K_{0}

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

. Then molecular mass of non-volatile substance is :-

Q. a) The boiling point of benzene is

353.23

1.80

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

90

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354.11

Calculate the molar mass of the solute. [

K_{b}

= 2.53

^{- 1}

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

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