250 mL of an aqueous solution of 6.3 g of oxalic acid dihydrate is diluted to 2 L. The normality of the resultant solution is:

0.07 N

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0.05 N

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0.50 N

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0.04 N

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Solution

The correct option is B 0.05 N
n-factor for oxalic acid dihydrate = 2
$Normality = \frac{number of gram equivalents}{total volume of solution in mL} = \frac{6.3 \times 2 \times 1000}{126 \times 250} = 0.4 N$

Now, the solution is diluted to 2L.
$N_{1} V_{1} = N_{2} V_{2} \Rightarrow (0.4 \times 250) = (N \times 2000)$
Normality of the final solution = 0.05 N

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250 mL of an aqueous solution of 6.3 g of oxalic acid dihydrate is diluted to 2 L. The normality of the resultant solution is:

The correct option is B 0.05 Nn-factor for oxalic acid dihydrate = 2 Normality=number of gram equivalentstotal volume of solution in mL=6.3×2×1000126×250=0.4 N Now, the solution is diluted to 2L. N1V1=N2V2⇒(0.4×250)=(N×2000) Normality of the final solution = 0.05 N