3.92 , g of ferrous ammonium sulfate crystals | vedclass

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(A) The equivalent weight of ferrous ammonium sulfate is $392 \, g/eq$. The normality of the salt solution is $N_1 = \frac{3.92}{392} 	imes \frac{100

Vedclass · Accepted Answer

$3.51$

Vedclass · Answer

(A) The equivalent weight of ferrous ammonium sulfate is $392 \, g/eq$. The normality of the salt solution is $N_1 = \frac{3.92}{392} 	imes \frac{1000}{100} = 0.1 \, N$. Using the titration formula $N_1V_1 = N_2V_2$ for $20 \, mL$ of salt solution and $18 \, mL$ of $KMnO_4$: $0.1 	imes 20 = N_2 	imes 18$. Normality of $KMnO_4$ solution $(N_2)$ $= \frac{2}{18} = \frac{1}{9} \, N$. The equivalent weight of $KMnO_4$ in acidic medium is $31.6 \, g/eq$. Strength of $KMnO_4$ in $g/L = 	ext{Normality} 	imes 	ext{Equivalent weight} = \frac{1}{9} 	imes 31.6 \approx 3.51 \, g/L$.

$3.92 \, g$ of ferrous ammonium sulfate crystals are dissolved in $100 \, mL$ of water. During titration,$20 \, mL$ of this solution is required for complete oxidation by $18 \, mL$ of $KMnO_4$ solution. Calculate the weight of $KMnO_4$ in $1 \, L$ of the solution in grams.

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