To make 50 mL solution of $K C l$ of 1 M, how many grams of $K C l$ are required to add to this solution?

1.00 g

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1.86 g

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3.73 g

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7.45 g

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1.49 g

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Solution

The correct option is D 3.73 g
50 ml of 1 M solution of KCl contains $0.05 \times = 0.05$ moles.

Weight = Moles $\times$ molar mass

Weight of KCl = $0.05 \times 74.5 = 3.73 g$

Hence, option C is correct.

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