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An aqueous solution of 6.3 g of oxalic acid dehydrate is made up to 250 ML. The volume of 0.1 N NAOH required to completely naturalise 10 ML of this solution is :
Equivalent mass of C2H2O4.2H2O = 126/2 = 63Normality of oxalic acid = 6.3/63 * 1000/250 = 0.4 NWe know that,N1V1 = N2V2Where N1 is normality of oxalic acidV1 is volume of oxalic acidN2 is normality of sodium hydroxideV2 is volume of sodium hydroxideTherefore,0.4 * 10 = 0.1 * V2V2 = 40 ml.Therefore 40 ml of NaOH is required to completely neutralize 10 ml of oxalic acid.
Equivalent mass of C2H2O4.2H2O = 126/2 = 63
Normality of oxalic acid = 6.3/63 * 1000/250 = 0.4 N
We know that,
N1V1 = N2V2
Where N1 is normality of oxalic acid
V1 is volume of oxalic acid
N2 is normality of sodium hydroxide
V2 is volume of sodium hydroxide
Therefore,
0.4 * 10 = 0.1 * V2
V2 = 40 ml.
Therefore 40 ml of NaOH is required to completely neutralize 10 ml of oxalic acid.
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