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The P^(H) of a sample of KOH solution is...

The `P^(H)` of a sample of `KOH` solution is `12.3979`. The weight of solid `KOH` of `70%` pure required to prepare `2.5` lit of this solution is

A

`3.5g`

B

5g

C

8 g

D

6 g

Text Solution

Verified by Experts

The correct Answer is:
B
`P^(OH)=14-12.3979=1.6021`
No of `[OH^(-)]=2.5xx10^(-2)N`
`:.N=(w)/(eq.wt)xx(1000)/(V(ml))`
`W=56.2.5xx2.5xx10^(-2)=3.5g` 70% pure KOH means, 100g impure sub `. 70g KOH`
`? " " to 3.5g KOH 5g`
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