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One third litre of $x \mathrm{MK}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is required to completely oxidise $2 \mathrm{~L}$ of $0.1 \mathrm{M}$ ferrous ammonium sulphate in acid medium. What Is $x$ ?
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Verified Answer
The correct answer is:
0.1
$x M \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7+6 \mathrm{FeSO}_4+7 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 7 \mathrm{H}_2 \mathrm{O}+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+3 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{K}_2 \mathrm{SO}_4$
$\mathrm{FeSO} \cdot 4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O} \text { is Mohr salt }(0.1 \mathrm{M})$
Given, $V_1$ (volume of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ ) $=\frac{1}{3} \mathrm{~L}$
Here, $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is required to completely oxidise ferrous ammonium sulphate $(0.1 \mathrm{M})$. Therefore, molarity of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=0.1 \mathrm{M}$
Thus, option (2) is correct.
$\mathrm{FeSO} \cdot 4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O} \text { is Mohr salt }(0.1 \mathrm{M})$
Given, $V_1$ (volume of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ ) $=\frac{1}{3} \mathrm{~L}$
Here, $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is required to completely oxidise ferrous ammonium sulphate $(0.1 \mathrm{M})$. Therefore, molarity of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=0.1 \mathrm{M}$
Thus, option (2) is correct.
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