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JEE Main levelSingle correct optionDifficulty 2 of 3so125

$200 mL$ of $0.50 M$ $NaCl$ is mixed with $300 mL$ of $0.20 M$ $NaCl$ (volumes additive). The molarity of the final solution is

A $0.35 M$
B $0.16 M$
C $0.53 M$
D $0.32 M$

Answer: D

Moles add:

0.200 \times 0.50 + 0.300 \times 0.20 = 0.100 + 0.060 = 0.160

mol in

0.500

L. So

M = 0.160/0.500 = 0.32

M. The simple average 0.35 M ignores the different volumes.

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