1.

How much volume of 0.40 M Na_(2)S_(2)O_(3) would be required to react with the I_(2) liberated by adding excess of KI of 50 mLof 0.20 M M CuSO_(4)

Answer»

12.5 mL
25 mL
50 mL
2.5 mL

Solution :m.eq of `Na_(2)S_(2)O_(3)` = m eq of `CuSO_(4)`
`0.4xxVxx1=50xx0.2xx1 " V=25ml"`


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