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Standard XII

Chemistry

Molarity

50 mL aqueous...

Question

50 mL aqueous solution of $F e S O_{4}$ required 12 mL of 0.2 M $K M n O_{4}$ in acidic medium for complete oxidation. Calculate the molarity of ferrous sulphate solution. The reaction is:
$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

0.024

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0.24

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2.4

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2.4 \times 10^{- 3}

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Solution

The correct option is B 0.24
$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} ⟶ M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$
Normality= Molarity $\times$ n-factor
$n - f a c t o r$ of $K M n O_{4} = 5$ $V_{1} = 50 m l; V_{2} = 12 m l; N_{2} = 0.2 M$
$n - f a c t o r$ of $F e S O_{4} = 2$
Now, $N_{1} V_{1} = N_{2} V_{2}$
$\Rightarrow N_{1} \times 50 = (0.2 \times 5) \times 12$
$\Rightarrow N_{1} \times 50 = 1.0 \times 12$
$\Rightarrow N_{1} = 0.24$
$∴$ Molarity= $0.24$

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50 mL aqueous solution of FeSO4 required 12 mL of 0.2 M KMnO4 in acidic medium for complete oxidation. Calculate the molarity of ferrous sulphate solution. The reaction is:MnO−4+5Fe2++8H+→Mn2++5Fe3++4H2O

The correct option is B 0.24MnO−4+5Fe2++8H+⟶Mn2++5Fe3++4H2ONormality= Molarity × n-factorn−factor of KMnO4=5 V1=50ml;V2=12ml;N2=0.2Mn−factor of FeSO4=2Now, N1V1=N2V2⇒N1×50=(0.2×5)×12⇒N1×50=1.0×12⇒N1=0.24∴ Molarity= 0.24

50 mL aqueous solution of $F e S O_{4}$ required 12 mL of 0.2 M $K M n O_{4}$ in acidic medium for complete oxidation. Calculate the molarity of ferrous sulphate solution. The reaction is:
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