A solution is prepared by dissolving 5.6 g of KOH per litre of solution. pOH of the solution would be

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$10^{- 0.1}$

$10^{- 1}$

answer is C.

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Detailed Solution

$[KOH] = \frac{5.6}{56} = 0.1 M = [{OH}^{-}] pOH = - \log [{OH}^{-}] = - \log (10^{- 1}) = 1$

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