Calculate the molecular mass of the caustic potash :

K_{2} O H

= 95 g m o l^{1}

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K O H = 56 g m o l^{1}

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K (O H)_{2}

= 73 g m o l^{1}

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K O H = 58 g m o l^{1}

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Solution

The correct option is B $K O H = 56 g m o l^{1}$
Potassium has valency $+ 1$ and is represented as $K^{+}$ , hydroxyl has valency $- 1$ and is represented as $O H^{-}$ . Caustic potash is $K O H .$ Molar mass - $(39 + 16 + 1) = 56 g m o l^{- 1}$

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