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The freezing point of a 0.05 m BaCl(2) i...

The freezing point of a 0.05 m `BaCl_(2)` in water ( 100% ionisation) is about `(K_(f)=1.86 Km^(-1))` :

A

`-0.279^(@)C`

B

`-0.558^(@)C`

C

`-0.093^(@)C`

D

`-0.186^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
A
`BaCl_(2) hArr Ba^(2)+2Cl^(-), i=3`
`DeltaT_(f)=iK_(f) xx m`
`DeltaT_(f)=3xx1.86 xx 0.05=0.279^(@)C`
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