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

Options

A. `3.5 g`
B. `5 g`
C. `8 g`
D. `6 g`

Answer

B

Explanation

`P^(OH) = 14 - 12.3979 = 1.6021`
`N` of `[OH^(-)] = 2.5 xx 10^(-2) N`
`:. N = (w)/("eq.wt') xx (1000)/(V(ml))`
`2.5 xx 10^(-2) = w/(56) xx (1)/(2.5(l))`
`w = 56 xx 2.5 xx 2.5 xx 10^(-2) = 3.5g`
`70%` pure KOH means,
`100 g` impure sub `rarr 70g KOH`
`? rarr 3.5g KOH`
`5g`