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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) = 39.7K "molality"^(-1))`, it lowered the freezing point by `0.62^(@)C`. The molar mass of eugenol is:

A

(a) `160g mol^(-1)`

B

(b) `165g mol^(-1)`

C

(c ) `200g mol^(-1)`

D

(d) `250g mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`m = (1000 xx K_(f) xx w)/(W xx Delta T)`
`= (1000 xx 39.7 xx 250 xx 10^(-3))/(100 xx 0.62)`
`= 160 g mol^(-1)`
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