How much water should be added in 3M, 100 ml $H_{2} S O_{4}$ solution to make it 2N solution?

100 mL

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300 mL

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200 mL

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Solution

The correct option is C 200 mL
Normality = molarity $\times$ n factor
$= 3 M \times 2 = 6 N$
Now apply dilution law :
$N_{1} V_{1} = N_{2} V_{2}$
$6 N \times 100 m l = 2 N \times V_{2}$
$V_{2} = \frac{6 N \times 100 m l}{2 N} = 300 m L$
Volume added = 300 – 100 = 200 ml

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