$3.458 g$ of Mohr's salt (molar mass $392 g mo l^{- 1}$ ) is dissolved in water and made up to $250.0 mL$ . A $25.0 mL$ portion is acidified and titrated with $0.0200 M$ $KMnO X_{4}$ . Find the volume of $KMnO X_{4}$ required, in $mL$ (two decimals).

Numerical answer type, answer in mL, tolerance 0.03.

Answer: 8.82 mL

Moles of Mohr's salt in the flask

= 3.458/392 = 8.821

mmol, so the

25.0 mL

portion has

0.8821

mmol of

Fe X^{2 +}

. Each mmol of

KMnO X_{4}

oxidises 5 mmol of

Fe X^{2 +}

: volume

= 0.8821/ (0.0200 \times 5) = 8.821 mL