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Question

The volume of 3 N HCl required to prepare 500 mL of an aqueous solution of 2 N HCl is:

A
750 mL
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B
150 mL
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C
333.33 mL
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D
233.33 mL
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Solution

The correct option is C 333.33 mL
Let the volume of 3 M HCl solution be V mL.
Since, normality becomes equal to molarity for HCl.
M1V1=M2V2(3×V)=(2×500)
V=333.33 mL
Volume of 3 M solution required = 333.33 mL

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