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When 250 mg of eugenol is added to 100 g...

When `250 mg` of eugenol is added to `100 g` of camphor `(k_(f)=39.7" molality"^(-1))`, it lowered the freezing point by `0.62^(@)C` . The molar of eugenol is `:`

A

`1.6xx10^(2) g//mol`

B

`1.6xx10^(4)g//mol`

C

`1.6xx10^(3)g//mol`

D

`200g//mol`

Text Solution

Verified by Experts

The correct Answer is:
1
`0.62=(250xx10^(-3))/(Mxx100)xx1000xx39.7xx1`
`M=160` or `1.6xx10^(2)g//mol`
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