JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 19)
In basic medium $$Cr{O_4}^{2 - }$$ oxidises $${S_2}{O_3}^{2 - }$$ to form $$S{O_4}^{2 - }$$ and itself changes into $$Cr{(OH)_4}^ - $$. The volume of 0.154 M $$Cr{O_4}^{2 - }$$ required to react with 40 mL of 0.25 M $${S_2}{O_3}^{2 - }$$ is __________ mL. (Rounded off to the nearest integer)
Answer
173
Explanation
$$17{H_2}O + 8Cr{O_4} + 3{S_2}{O_3}\buildrel {} \over
\longrightarrow 6S{O_4} + 8Cr{(OH)_4}^ - + 2O{H^ - }$$
Applying mole-mole analysis
$${{0.154 \times v} \over 8} = {{40 \times 0.25} \over 3}$$
v = 173 mL
Applying mole-mole analysis
$${{0.154 \times v} \over 8} = {{40 \times 0.25} \over 3}$$
v = 173 mL
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