The pH 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
3.5 g
5 g
8 g
6 g
POH = 14 - 12.3979 ~ 1.6
[OH-] = 2.5 x10-2
Let us consider, w g of KOH is required
70×w100×56×2.5=2.5×10-2
⇒ w = 5