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 (integer)
173
Solution
$$17{H_2}O + 8Cr{O_4} + 3{S_2}{O_3}\buildrel {} \over
\longrightarrow 6S{O_4} + 8Cr{(OH)_4}^ - + 2O{H^ - }$$<br><br>Applying mole-mole analysis<br><br>${{0.154 \times v} \over 8} = {{40 \times 0.25} \over 3}$<br><br>v = 173 mL
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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