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Question

If $X %$ of $C r$ in a sample of dichromate ore if $1.0 g$ of the sample after fusion is treated with $60 m L$ of $0.1 N$ $F e S O_{4} \cdot (N H_{4})_{2} S O_{4}$ and the excess of $F e^{2 +}$ requires $11.2 m L$ of $K_{2} C r_{2} O_{7} (1 m L$ of $K_{2} C r_{2} O_{7} = 0.006 g F e)$ . Also find $%$ of $C r_{2} O_{3}$ in the sample then calculate the value of $100 X$ .

1216

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1200

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1250

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Solution

The correct option is A 1216
$6 e^{-} + C r_{2} O_{7}^{2 -} \to 2 C r^{3 +} (n = 6)$

$F e^{2 +} \to F e^{3 +} + e^{-} (n = 1)$

mEq of $K_{2} C r_{2} O_{7}$ in $1 m L$ = mEq of $F e$

$= \frac{0.006}{56} \times 10^{3} = \frac{6}{56}$

mEq of $K_{2} C r_{2} O_{7}$ in $11.2 m L = \frac{6}{56} \times 11.2 = 1.2$

mEq of $F e^{2 +}$ left unused=mEq of $K_{2} C r_{2} O_{7}$ used $= 1.2$

Now, mEq of $F e S O_{4} \cdot (N H_{4})_{2} S O_{4}$ added $= 60 \times 0.1 = 6$

mEq of $F e S O_{4} \cdot (N H_{4})_{2} S O_{4}$ left unused $= 6 - 1.2 = 4.8$

mEq of $C r = 4.8$ $(∵ E w o f C r = 52 / 3)$

$\frac{W}{52 / 3} \times 10^{3} = 4.8$

$W_{C r} = \frac{4.8 \times 52}{10^{3} \times 3} = 0.0832 g$

$%$ of $C r = \frac{0.0832}{1.0} \times 100 = 8.32 %$

Also, mEq of $C r_{2} O_{3} = 4.8$

$(∵ E w o f C r_{2} O_{3} = \frac{52 \times 2 + 48}{6} = \frac{156}{6})$

$\frac{W_{C r_{2} O_{3}}}{152 / 6} \times 10^{3} = 4.8$

$W_{C r_{2} O_{3}} = \frac{4.8 \times 152}{6 \times 1000} = 0.1216 g$

$%$ of $C r_{2} O_{3} = \frac{0.1216}{1.0} \times 100 = 12.16 %$

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If X% of Cr in a sample of dichromate ore if 1.0 g of the sample after fusion is treated with 60 mL of 0.1N FeSO4⋅(NH4)2SO4 and the excess of Fe2+ requires 11.2 mL of K2Cr2O7(1 mL of K2Cr2O7=0.006gFe). Also find % of Cr2O3 in the sample then calculate the value of 100X.