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CGP EDU Academic Team
Published on: August 14, 2026
50 ml of 0.2 M solution of a compound with empirical formula CoCl 3 .4NH 3 on treatment with excess of AgNO 3 (aq) yields 1.435 g of AgCl. Ammonia is not removed by treatment with concentrated H 2 SO 4. The formula of the compound is:
Text Solution
Verified by ExpertsThe correct answer is:
B
mole of complex = 50 × 0.2 = 0.01 and mole of AgCl = 
= 0.01
nAg + = nCl –
∴ 1 mole complex = 1 mole AgCl
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