1.96 g of Mohr's salt, FeSO4.(NH4)2SO4.6H2O (molar mass 392), is dissolved in dilute H2SO4 and titrated with 0.02 M KMnO4. The volume of KMnO4 required is:

JEE Main levelSingle correct optionDifficulty 2 of 3rx178

1.96 g of Mohr's salt, $FeSO X_{4} \cdot (NH X_{4}) X_{2} SO X_{4} \cdot 6 H X_{2} O$ (molar mass 392), is dissolved in dilute $H X_{2} SO X_{4}$ and titrated with 0.02 M $KMnO X_{4}$ . The volume of $KMnO X_{4}$ required is:

Moles of Mohr's salt = 1.96/392 = 5 mmol, giving 5 meq of

Fe X^{2 +}

(n = 1). Normality of

KMnO X_{4}

= 0.10 N, so

V = 5/0.10 = 50

mL.