Please help with this Chem question!

Question

It takes 32.47 mL of a KMnO4 solution to neutralize 2.380 g of sodium oxalate. What is the normality of the KMnO4 solution? The EW of Na2C2O4 is 67.0 g/eq.

2) It takes 37.32 mL of a 0.103 N KMnO4 solution to neutralize 23.53 mL of an unknown oxalic acid solution. What is the normality of the oxalic acid solution?

I really don't know how to do this. Please help!

Answer 1

equivalents KMnO4 = equivalents Na2C2O4

(0.03247litres)  (N)  =  (2.380 grams)(1 equiv / 67.0 grams)

N = 0.035522 equiv / 0.03247 litres

first answer = 1.094 N

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equivalents KMnO4 = equivalents Na2C2O4

(0.03732litres) (0.103 equiv/litre)  =  (0.02353 litres)(N)

N = 0.003844 equiv / 0.02353 litres

next answer: 0.163 N