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Question

1.22 g of a monobasic acid is dissolved in 100 g of benzene. Boiling point of solution increases by 0.13C with respect to pure benzene. Find the molecular mass of acid in benzene solvent (in u). Report your answer after dividing it by 100 and round it off to the nearest integer.

(Kb of benzene=2.6 K kg mol1)

A
2.0
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B
2.00
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C
2
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Solution

Given,
B.P elevation =0.13
Here, m = Mass of benzene=100 g=0.1 kg
Kb of benzene=2.6 K kg mol1=2.6C kg mol1

ΔT=i Kbm .....(i)

Where ΔT is elevatioin in boiling point

Kb is molal elevation constant

i is van't hoff factor (i for benzene=1)

m is molality

Molality=moles of solutemass of solvent (kg)

Let the number of moles of acid in benzene = x
Equation (i) becomes :

0.13C=(1)(2.6C kg mol1)×x0.1 kg

0.13=2.6×x0.1

x=1/200 moles
x=0.005 moles

moles=Given massMolar mass

0.005=1.22 gmolecular weight

molecular weight=1.22 g0.005 moles

Molecular weight of acid=244100=2.44 u2 u

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