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Byju's Answer

Standard XII

Chemistry

EMF

0.2828 g of i...

Question

0.2828 g of iron wire was dissolved in excess of dilute $H_{2} S O_{4}$ and the solution was made upto 100 ml. 20 ml of this solution required 30 ml. of $\frac{N}{30} K_{2} C r_{2} O_{7}$ solution for oxidation. Calculate % purity of iron in the wire:

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Solution

The correct option is A 99
$F e + H_{2} {S O}_{4} ⟶ {F e S O}_{4} + H_{2}$
Millequivalents of $F e =$ Milliequivalents of
FeSO $_{4}$
$=$ Milliequivalets for second reaction
$k_{2} {C r}_{2} O_{7}$
$\begin{matrix} N_{1} v_{1} & = N_{2} V_{2} M \times \frac{1000}{100} \times 20 = \frac{N}{30} \times 30 & M = moles of Iron M = \frac{1}{200} amount of iron in sample & = \frac{56}{200} = 0.28 \end{matrix}$
$\begin{matrix} So, percentage purity & = \frac{0.28}{0.2828} \times 100 = 99 % \end{matrix}$
$Hence option (A) 99% is correct.$

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0.2828 g of iron wire was dissolved in excess of dilute H2SO4 and the solution was made upto 100 ml. 20 ml of this solution required 30 ml. of N30K2Cr2O7 solution for oxidation. Calculate % purity of iron in the wire:

0.2828 g of iron wire was dissolved in excess of dilute $H_{2} S O_{4}$ and the solution was made upto 100 ml. 20 ml of this solution required 30 ml. of $\frac{N}{30} K_{2} C r_{2} O_{7}$ solution for oxidation. Calculate % purity of iron in the wire: