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JEE MAIN 2022
27-07-2022 S1
Question
20 mL of $0.02 \mathrm{M} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution is used for the titration of 10 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)
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Solution
Eq. of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=$ Eq. of $\mathrm{Fe}^{2+}$ $$ \begin{array}{ll} \Rightarrow \quad & \text { (Molarity × volume × n.f) of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=(\text { molarity } \times \text { volume } \times \mathrm{n} . \mathrm{f}) \text { of } \mathrm{Fe}^{2+} \\ \Rightarrow \quad & 0.02 \times 20 \times 6=\mathrm{M} \times 10 \times 1 \\ \Rightarrow \quad & \mathrm{M}=0.24 \\ \Rightarrow \quad & \text { Molarity }=24 \times 10^{-2} \end{array} $$
Question Tags
JEE Main
Chemistry
Easy
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