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Question

An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is :


A
40 mL
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B
20 mL
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C
10 mL
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D
4 mL
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Solution

The correct option is A 40 mL
n-factor for $$H_2C_2O_4. 2H_2O$$=2

Normality =$$\dfrac{weight \times 2 \times 1000}{molecular \ weight \times 250}$$ = $$\dfrac{ 6.3 \times 2000}{126 \times 250}$$ =0.4

equating no. of equivalents :

$$N_1V_1=N_2V_2$$

$$0.4\times 10= V_2  \times 0.1$$

Volume ($$V_2$$)= 40mL.

Hence, the correct option is $$A$$

Chemistry

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