3-92 g - L of a sample of ferrous ammonium sulphate F eSO 4- NH 42 SO 4 -6 H 2 O reacts completely with 50 mL N -10 KMnO 4 solution- Thepercentage purity of the sample is -A- 50 - B- 39-2 - 78-4 -D- 80-0 -

Name: NEET - Grade 11 - Chemistry - Hydrogen and the S-block - Session 01 - W02
Uploaded: 2022-07-04T10:36:42+05:30
Description: The correct option is A 50 %Ferrous ammonium sulphate ≡ FeSO_4.(NH_4)_2 SO_4 . 6H_2O Equivalents of FeSO_4 . (NH_4)_2 SO_4 . 6H_2O = Equivalents of KMnO_4 N_1V ...

The correct option is A 50 %Ferrous ammonium sulphate ≡ FeSO_4.(NH_4)_2 SO_4 . 6H_2O Equivalents of FeSO_4 . (NH_4)_2 SO_4 . 6H_2O = Equivalents of KMnO_4 N_1V ...

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Byju's Answer

Standard XII

Chemistry

Equivalent Mass

3.92 g / L of...

Question

$3.92 g / L$ of a sample of ferrous ammonium sulphate $(F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O)$ reacts completely with $50 m L \frac{N}{10} K M n O_{4}$ solution. The percentage purity of the sample is :

50 %

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78.4 %

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80.0 %

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39.2 %

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Solution

The correct option is A $50 %$
$Ferrous ammonium sulphate \equiv F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O$
$Equivalents of F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O = Equivalents of K M n O_{4}$
$N_{1} V_{1} = N_{2} V_{2}$
$N_{1} \times 1000 = \frac{1}{10} \times 50$
$N_{1} = \frac{1}{200}$
$Equivalent weight of F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O = Molecular weight \times 1 = 392$

$∴ Strength of pure salt = 392 \times \frac{1}{200} = 1.96 g / L$
$∴$ % purity $= \frac{1.96}{3.92} \times 100 = 50 %$

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3.92 g / L

of a sample of ferrous ammonium sulphate

(F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O)

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50 m L \frac{N}{10} K M n O_{4}

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Q. A sample of hydrazine sulphate

(N_{2} H_{6} S O_{4})

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mL of water and

10

mL of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required

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mL of

\frac{M}{50}

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4 F e^{3 +} + N_{2} H_{4} \to N_{2} + 4 F e^{2 +} + 4 H^{⨁}

M n O_{4}^{⊝} + 5 F e^{2 +} + 8 H^{⨁} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

A sample of hyhdrazine sulphate $(N_{2} H_{6} S O_{4})$ was dissolved in 100 ml water. 10 ml of this solution was reacted with excess of solution and warmed to complete the reaction. Ferrous ions formed required 20 ml of solution.
Given : $4 F e^{3 +} + N_{2} H_{4} \to N_{2} + F e^{2 +} + 4 H^{+}$
$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

The amount (g) of hydrazine sulphate in one litre is

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Equivalent Mass

Standard XII Chemistry

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3.92 g/L of a sample of ferrous ammonium sulphate (FeSO4.(NH4)2SO4.6H2O) reacts completely with 50 mL N10KMnO4 solution. The percentage purity of the sample is :

$3.92 g / L$ of a sample of ferrous ammonium sulphate $(F e S O_{4} . (N H_{4})_{2} S O_{4} . 6 H_{2} O)$ reacts completely with $50 m L \frac{N}{10} K M n O_{4}$ solution. The percentage purity of the sample is :