JEE MAIN - Chemistry (2022 - 27th July Morning Shift - No. 14)
$$20 \mathrm{~mL}$$ of $$0.02 \,\mathrm{M} \,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$$ solution is used for the titration of $$10 \mathrm{~mL}$$ of $$\mathrm{Fe}^{2+}$$ solution in the acidic medium.
The molarity of $$\mathrm{Fe}^{2+}$$ solution is __________ $$\times \,10^{-2}\, \mathrm{M}$$. (Nearest Integer)
Answer
24
Explanation
Applying the law of equivalence, milliequivalents of $\mathrm{Fe}^{2+}=$ milliequivalents of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
$$ \begin{aligned} &10 \times 1 \times M=20 \times 6 \times .02 \\\\ &M=24 \times 10^{-2} M \end{aligned} $$
$\therefore $ Answer will be 24
$$ \begin{aligned} &10 \times 1 \times M=20 \times 6 \times .02 \\\\ &M=24 \times 10^{-2} M \end{aligned} $$
$\therefore $ Answer will be 24
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