You visited us 1 times! Enjoying our articles? Unlock Full Access!

Packing Fraction and Efficiency

Volume of 0...

Question

Volume of $0.02$ M ${M n O_{4}}^{-}$ solution required to oxidise $40.0$ mL of $0.1$ M $F e^{2 +}$ solution is :

200 mL

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 mL

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 mL

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

20 mL

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 40 mL
${M n O_{4}}^{-} + 5 F e^{2 +} \to 5 F e^{3 +} + M n^{2 +}$
$M_{1} V_{1} ({M n O_{4}}^{-}) = 5 M_{1} V_{1} ({M n O_{4}}^{-})$ milliequivalents
$M_{2} V_{2} (F e^{2 +}) = M_{2} V_{2} (F e^{2 +})$ milliequivalents
$N_{1} V_{1} ({M n O_{4}}^{-}) = N_{2} V_{2} (F e^{2 +})$
$5 \times 0.02 \times V_{1} = 40 \times 0.1$
$V_{1}$ $= 40$ mL.

Suggest Corrections

Similar questions

F e^{2 +}

H N O_{2}

M n O_{4}^{-}

F e^{2 +}

V

M n O_{4}^{-} / H^{+}

H N O_{2}

V

M n O_{4}^{-} / H^{+}

0.1 M K_{2} C r_{2} O_{7}

35 m L

0.5 M F e S O_{4}

0.50 M

N a O H

40.0

0.05

H_{2} S O_{4}

K M n O_{4}

Join BYJU'S Learning Program