3.92 g of ferrous ammonium sulphate react completely with 50 ml (N/10) KMnO4 solution. The percentage purity of the sample is

Question

3.92 g of ferrous ammonium sulphate react completely with 50 ml (N/10) KMnO4 solution. The percentage purity of the sample is (A) 50 (B) 78.4 (C) 80 (D) 39.2

Tardigrade Admin · Accepted Answer

Eq of KMnO 4 used =(50 × 1/1000 × 10)=0.005 ∴ Eq of FAS reacted =0.005 ∴ weight of FAS needed =0.005 × 392=1.96 g Thus percentage purity of FAS is 50 %

Q. $3.92 g$ of ferrous ammonium sulphate react completely with $50 m l \frac{N}{10} K M n O_{4}$ solution. The percentage purity of the sample is

50

78.4

80

39.2

$Eq$ of $K M n O_{4}$ used $= \frac{50 \times 1}{1000 \times 10} = 0.005$
$∴$ Eq of FAS reacted $= 0.005$
$∴$ weight of FAS needed
$= 0.005 \times 392 = 1.96 g$
Thus percentage purity of FAS is $50%$

Q. 3.92g of ferrous ammonium sulphate react completely with 50ml10N​KMnO4​ solution. The percentage purity of the sample is

50

78.4

80

39.2

Eq of KMnO4​ used =1000×1050×1​=0.005 ∴ Eq of FAS reacted =0.005 ∴ weight of FAS needed =0.005×392=1.96g Thus percentage purity of FAS is 50%

Q. $3.92 g$ of ferrous ammonium sulphate react completely with $50 m l \frac{N}{10} K M n O_{4}$ solution. The percentage purity of the sample is

$Eq$ of $K M n O_{4}$ used $= \frac{50 \times 1}{1000 \times 10} = 0.005$
$∴$ Eq of FAS reacted $= 0.005$
$∴$ weight of FAS needed
$= 0.005 \times 392 = 1.96 g$
Thus percentage purity of FAS is $50%$