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In an experiment, 50 mL of 0.1 M solution of a metallic salt reacted exactly with 25 mL of 0.1 M solution of sodium sulphite. In the reaction, SO3 is oxidised to SO42. If the original oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?

Solution
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SO32 is oxidised to SO42 (change in O.N =2)
25 mL of 0.1 M SO32 =2.5 millimoles =5.0 milliequivalents of SO32=5.0 milliequivalents of M3+
50 mL of 0.1 M M3+ should be 1.
So, 5 millimoles = 5 milliequivalents. Thus, new O.N of metal = 2 as the metal is reduced by 1.

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In an experiment, 50 mL of 0.1 M solution of a metallic salt reacted exactly with 25 mL of 0.1 M solution of sodium sulphite. In the reaction, SO3 is oxidised to SO42. If the original oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?
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