an aqeous sollution of 6.3 g oxalic acid is made up to 250 mL. calculate the volume of 0.1 N sodium hydroxide needed to neutralise 10 mL of this sollution?

Answer 1

As per previous question, the stoichiometry gives:
m/w = cV/2

However, this time your given the mass of oxalic acid, in the aloquat, and wish to find V.

Given:
m = (6.3)/25 g
c = 0.1N = 0.1M
w = 90.03517 g/mol
V = 2m/cw

Answer 2

40 mL