question_answer

How many mL of\[0.125\text{ }M\text{ }C{{r}^{3+}}\]must be reacted with 12.0 mL of\[0.200\text{ }M\text{ }MnO_{4}^{-}\] if the redox products are\[C{{r}_{2}}O_{7}^{2-}\]and\[M{{n}^{2+}}\]?

A)  32 mL  

B)  24 mL  

C)  16 mL 

D)  8 mL

Correct Answer: A

Solution :

  Equivalents of\[C{{r}^{3+}}=3\times \]moles of\[C{{r}^{3+}}\] Equivalents of \[MnO_{4}^{-}=5\times \]moles of\[MnO_{4}^{-}\] Amount of \[C{{r}^{3+}}=0.125\times V\,\]milli mol \[=0.125\times V\times 3\,\]miliequiv. Amount of\[MnO_{4}^{-}=0.200\times 12.00\times 5\]milliequiv \[\therefore \]\[0.125\times V\times 3=0.200\times 12.00\times 5\] \[V=32.0\,mL\]