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Addition of 0.643 g of compound to 50 ml...

Addition of 0.643 g of compound to 50 ml of liquid (density 0.879 g/ml) lowers the freezing point from `5.51^@C` to `5.03^@C`. Calculate the molar mass of the compound. [`K_f` for benzene =5.12 K Kg `mol^(-1)`]

Text Solution

Verified by Experts

The correct Answer is:
`Mw=156.06 g mol^(-1)`
According to depression in freezing point method
`Mw_(2)=(1000xxK_(f))/(DeltaT_(f)) xx W_(2)/W_(1)`
Given that
`K_(f) rarr` molar depression constant of benzene
`=5-12 K kg "mol"^(-1)`
`W_(2)` =Weight of compound =0.643 g
`W_(1)` =Weight of benzene =volume`xx` Density
`=50 xx 0.879`
`=43.95 g`
`DeltaT_(f) rarr` Depression in freezing point =`5.51-5.03=0.48^(@)C`
Therefore, on substitution,
`Mw_(2)=(1000xx5.12xx0.643)/(0.48xx43.95)=156.06`
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