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Standard XII

Chemistry

Elevation in Boiling Point

Y g of non-vo...

Question

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 :

\frac{M}{K_{b} Y}

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\frac{4 K_{b} Y}{M}

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\frac{K_{b} Y}{4 M}

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\frac{K_{b} Y}{M}

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Solution

The correct option is B $\frac{4 K_{b} Y}{M}$
Elevation in its boiling point is given by

$Δ T = \frac{1000 \times K_{b} \times Y}{250 \times M}$

$=$ $\frac{4 K_{b} Y}{M}$

Hence, the correct option is $B$

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Similar questions

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:

Q. A solution of

2.5 g

of a non - volatile solid in

100 g

benzene is boiled at

0.42 K

higher than the boiling point of pure benzene. Calculate the molar mass of the substance. Molal elevation constant of benzene is

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

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 .

Q. Boiling point of benzene is

353.23

K. When

1.8

g of non-volatile solute is dissolved in

90

g of benzene. Then boiling point raised to

354.11

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K_{0}

(benzene) =

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

. Then molecular mass of non-volatile substance 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. $(K_{b} f o r b e n z e n e i s 2.53 K k g m o l^{- 1})$

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Y g of non-volatile organic substance of molar mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is Kb. Elevation in its boiling point is given by :

The correct option is B 4KbYMElevation in its boiling point is given by ΔT=1000×Kb×Y250×M =4KbYMHence, the correct option is B

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 :

The correct option is B $\frac{4 K_{b} Y}{M}$
Elevation in its boiling point is given by

$Δ T = \frac{1000 \times K_{b} \times Y}{250 \times M}$

$=$ $\frac{4 K_{b} Y}{M}$

Hence, the correct option is $B$