Question

25 ml of a solution containing 6.1 g litre^-1 of an oxalate of formula К_аH_b (C₂O₄)с.nН₂O required for titration 18 ml of 0.1 N NaOH and 24 ml of 0.1 N KMnO₄ for oxidation. Calculate a, b, c and n.

Answer 1

Normality of oxalate as an acid = \(\frac{18 \times 0.1}{25} = 0.072\)

\(\therefore\) Eq. weight as an acid = \(\frac{6.1}{0.072} = 84.72\)

Normality of oxalate as a reducing agent = \(\frac{24 \times 0.1}{25} = 0.096\)

\(\therefore\) Eq. weight as a reducing agent = \(\frac{6.1}{0.096}\) = 63.54

Now we can write

\(\frac{\text{Molecular weight}}b = 84.72\) and \(\frac{\text{Mol. wt.}}{2c} = 63.54\)

b = 1.5c but (a + b) = 2c

\(\therefore\) a, b and c are in ratio of 1 : 3 : 2

or a = 1, b = 3 and c = 2

\(\therefore\) Mol. wt. of oxalate = 84.72 x 3 = 254

and 39a + b + 88c + 18n = 254

18n = 36

n = 2