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The molal freezing point depression cons...

The molal freezing point depression constant for benzene is `4.9 K kg mol^(-1)`. Selenium exists as a polymer of the type `Se_(n)`. When `3.26 g` selenium is dissolved in `226 g` of benzene, the observed freezing. If molecular formula of sulphur is `S_(n)`. Then find the value of `n`. (at. wt. of `S = 32`).

Text Solution

Verified by Experts

The correct Answer is:
8
`Delta T_(f) = (K_(f) xx w xx 1000)/(m xx W)` Given, `Delta T_(f) = 0.112 K, K_(f) = 4.9, w = 3.26g`,
`W = 226g`
`:. m = (4.9 xx 3.26 xx 1000)/(0.112 xx 226) = 631.08`
Now mol. Mass of `Se = 78.8`
`because n xx 78.8 = 631.08`
`:. n = 8`
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