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Q.

How many milli-Iitre of 0.5 N SnCl2 solution will reduce 600 mL of 0.1 N HgCl2 to Hg2Cl2?

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a

120 mL

b

60 mL

c

30 mL

d

240 mL

answer is A.

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Detailed Solution

SnCl2+2+Hg+2Cl2⟶Hg2+1Cl2+Sn +4
0.5 x V(mL) = 600 x0.1
V(mL) =120 mL

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