Q. Volume of 0.1 M $K_{2}{Cr}_{2}O_{7}$ required to oxidise 35 mL of 0.5 M $Fe{SO}_{4}$ solution is
A) 29.2 mL
B) 17.5 mL
C) 175 mL
D) 145 mL
Explanation
(a)
${Cr}_{2}O_{7}^{2 -}\ + \ 6{Fe}^{2 +} + 14H^{+}\ \longrightarrow 6Fe^{3 +} + \ {2Cr}^{3 +} + \ 7H_{2}O$ Hence, 1 mol of ${Cr}_{2}O_{7}^{2 -}$ = 6 moles of $Fe^{2 +}$
$\frac{M_{1}V_{1}}{1} = \frac{M_{2}V_{2}}{6}$
$\frac{0.1{\times V}_{1}}{1} = \frac{0.5\ \times \ 35}{6}$
$V_{1} = \ \ \frac{0.5\ \times \ 35}{6\ \times \ 0.1}$
${\ \ V}_{1}$ = 29.2 mL
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.