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To 500 cm^(3) of water, 3.0xx10^(-3) kg ...

To `500 cm^(3)` of water, `3.0xx10^(-3) kg` of acetic acid is added. If `23%` of acetic acid is dissociated, what will be the depression in freezing point? `K_(f)` and density of water are `1.86 K kg^(-1) mol_(-1)` and `0.997 g cm^(-3)`, respectively.

A

`0.186K`

B

`0.228K`

C

`0.372K`

D

`0.556K`

Text Solution

Verified by Experts

The correct Answer is:
B
Weight of water `=500xx0.997=498.5g`
No. of moles of acetic acid =("Wt. of" CH_(3)COOH("in" gm))/("mol.wt.of" CH_(3)COOH)=(3xx10^(-3)xx10^(3))/(60)=0.05`
Since `498.5g` of water has `0.05` moles of `CH_(3)COOH`
`1000g` of water has `=(0.05xx1000)/(498.5)=0.1`
Determination of van't Hoff factor, `i`
`CH_(3)COOHrarrCH_(3)COO^(-)+H^(+)`
`{:("No of moles at start",1,0,0),("No. of moles at equb.",1-0.23,0.23,0.23):}`
`DeltaT_(f)=(1-0.23+0.23+0.23)xx1.86xx0.1=0.228K`.
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