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The elevation in boiling point would be ...

The elevation in boiling point would be highest for

A

`0.08 m BaCl_(2)`

B

`0.10 m` glucose

C

`0.10 m KCl`

D

`0.06 m` calcium nitrate

Text Solution

Verified by Experts

The correct Answer is:
a
More the number of particles and concentration higher is the elevation in boiling point. Thus, 0.08 m KCl
`= -0.082xx2 = 0.24m` (highest)
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