Redox and Equivalent ConceptHard
Question
The volume of 0.10 M-AgNO3 should be added to 10.0 ml of 0.09 M-K2CrO4 to precipitate all the chromate as Ag2CrO4 is
Options
A.18 ml
B.9 ml
C.27 ml
D.36 ml
Solution
$n_{eq}AgNO_{3} = n_{eq}K_{2}CrO_{4}$
Or $\frac{V \times 0.1}{1000} \times 1 = \frac{10 \times 0.09}{1000} \times 2 \Rightarrow V = 18\text{ ml}$
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