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Q.

4I⊖+Hg2+→HgI42−; 1 mole each of Hg2+ and I⊖

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a

1 mol of HgI42-

b

0.5 mol of HgI42-

c

0.25 mol of HgI42-

d

2 mol of HgI42-

answer is C.

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Detailed Solution

Hg2+    ≡ 4I⊖    =HgI42−1mol    ≡ 4mol    ≡    1mol14mol    ≡44mol    =14mol    ≡0.25mol
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