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The addition of 0.643 g of a compound to...

The addition of `0.643 g` of a compound to `50 mL` of benzene (density 0.879 g `mL^(-1)`) lowers the freezing point from `5.51` to `5.03^(@)C`. If `K_(f)` for benzene is `5.12`, calculate the molecular weight of the compound.

Text Solution

Verified by Experts

According to depression in freezing point method
`m=(1000xxk_f)/(DeltaT_f)xxw/W`
Given that,
`k_f to ` molar depression constant of benzene
`=5.12 K kg mol ^(-1)`
`w to ` weight of compound = 0.643 g `W to ` weight of benzene = volume density
`= 50 xx0.879 = 43.95 g`
`D T_f to ` depression in freezing point
`=(5.51 -5.03 )=0.48^@C`
`therefore` On substitution
`m= (1000 xx5.12xx0.643)/(0.48xx43.95)=156.06`
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