Question

0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium

Answer 1

Moles of potassium permanganate = 0.0008

Further explanation  

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range  

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) +  10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium  permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002$

From equation, mol ratio  Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

$\tt \dfrac{2}{5}\times 0.002=0.0008$