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1.00 g of a non-electrolyte solute (mola...

1.00 g of a non-electrolyte solute (molar mass 250g `mol^(–1)`) was dissolved in 51.2 g of benzene. If the freezing point depression constant `K_(f)` of benzene is 5.12 K kg `mol^(–1)`, the freezing point of benzene will be lowered by:-

A

0.4 K

B

0.3 K

C

0.5 K

D

0.2 K

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaT_(f)=molalityxxK_(f)`
`DeltaT_(f)=((1xx1000)/(250xx51.2))xx5.12`
`DeltaT_(f)=0.4K`
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1.0 g of a non-electrolyte solute( mol. Mass 250.0 g mol^(-1) ) was dissolved in 5.12 g benzene. If the freezing point depression constant, K_(f) of benzene is 5.12 K kg mol^(-1) , the freezing point of benzene will be lowered by:

1.0 g of a non-electrolyte solute (molar mass 250gmol^(-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is 5.12K kgmol^(-1) the lowering in freezing point will be

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