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Standard XII

Chemistry

Elevation in Boiling Point

Boiling point...

Question

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 $354.11$ K. Given $K_{0}$ (benzene) = $2.53 K k g m o l^{- 1}$ . Then molecular mass of non-volatile substance is :-

58 g m o l^{- 1}

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120 g m o l^{- 1}

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116 g m o l^{- 1}

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60 g m o l^{- 1}

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Solution

The correct option is A $58 g m o l^{- 1}$
$Δ T_{b} = 364.11 - 353.23$
$= 0.88 K$
$K_{b} = 2.53 K k g m o l^{-} 1$
$Δ T_{b} = \frac{K_{b} \times W_{2} \times 1000}{m_{2} \times W_{1}}$
$m_{2} = \frac{2.53 \times 1.8 \times 1000}{0.88 \times 90}$
$= 57.5 g$
So mass of non volatile solute is $= 57.5$ or $58$ gram.

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Similar questions

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. When

.15

g of a non-volatile solute was dissolved in

90

g of benzene, the boiling point of benzene raised from

353.23

K to

353.93

K. Calculate the molar mass of the solute.
(

K_{b}

for benzene

= 2.52

m o l^{- 1}

Q. Define the following
(a) Leaching
(b) Metallurgy
(c) Anisotropy
Derive an expression for maximum work.
The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute.
[

K_{b}

for benzene = 2.53 K mol

^{- 1}

]

Q. 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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Boiling point of benzene is 353.23K. When 1.8g of non-volatile solute is dissolved in 90g of benzene. Then boiling point raised to 354.11K. Given K0 (benzene) = 2.53Kkgmol−1. Then molecular mass of non-volatile substance is :-

The correct option is A 58gmol−1ΔTb=364.11−353.23 =0.88KKb=2.53Kkgmol−1ΔTb=Kb×W2×1000m2×W1m2=2.53×1.8×10000.88×90 =57.5gSo mass of non volatile solute is =57.5 or 58 gram.

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 $354.11$ K. Given $K_{0}$ (benzene) = $2.53 K k g m o l^{- 1}$ . Then molecular mass of non-volatile substance is :-