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Question

A solution containing 1.8 g of a non-electrolyte (empirical formula CH2O) in 40 g of water is observed to freeze at 0.465C. The molecular formula of the compound is:
(Kf of water =1.86 kg K mol1)

A
C2H4O2
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B
C3H6O3
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C
C4H8O4
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D
C6H12O6
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Solution

The correct option is D C6H12O6

We know,
ΔTf=Kf×molality
ΔTf=Depression of freezing point=0.465C
Kf=1.86 kg K mol1

0.465=1.86×1.8M×40×1000
M=1.86×1.8×10000.465×40=180 g mol1

Empircal formula given = CH2O
Let, molecular formula = (CH2O)n
Empirical formula mass =30 u
Molecular formula mass =180 u
n=Molecular formula massEmpirical formula mass=18030=6
So, molecular formula = C6H12O6.

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