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Molarity of a solution prepared by dissolving $75.5 g$ of pure $KOH$ in $540 ml$ solution is

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$3.05 M$

$1.35 M$

$2.50 M$

$4.50 M$

answer is C.

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Detailed Solution

$M = \frac{w \times 1000}{m \cdot w t \cdot \times V in m l} = \frac{75.5 \times 1000}{56 \times 540} = 2.50 M .$

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