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To prepare 0.1M KmnO(4) solution in 250m...

To prepare `0.1M KmnO_(4)` solution in 250mL flask, the weight of `KmnO_(4)` required is:

A

`4.80g`

B

`3.95g`

C

`39.5g`

D

`0.48g`

Text Solution

Verified by Experts

The correct Answer is:
B
`0.1 xx (n)/(250) xx 100 rArr n = (0.1)/(4)`
wt `KMnO_(4) = (0.1)/(4) xx 158 = 3.95g`
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