50 mL of 0.2 M solution of a compound with empirical formula CoCl3.4NH3 on treatment with excess of AgNO3(aq) yields 1.435g of AgCl. Ammonia is not removed by treatment with concentrated H2SO4. How many ionisable chlorine atoms are present in the formula of the compound?

Question

50 mL of 0.2 M solution of a compound with empirical formula CoCl3.4NH3 on treatment with excess of AgNO3(aq) yields 1.435g of AgCl. Ammonia is not removed by treatment with concentrated H2SO4. How many ionisable chlorine atoms are present in the formula of the compound? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Moles of AgCl = (1.435/143.5) = 0.01 ;50 mL of 0.2 M complex = 0.01 mol [Co(NH3)4 Cl2] Cl→ [Co(NH3)4 Cl2]+ +Cl-

Q. 50mL of 0.2M solution of a compound with empirical formula CoCl3​.4NH3​ on treatment with excess of AgNO3​(aq) yields 1.435g of AgCl. Ammonia is not removed by treatment with concentrated H2​SO4​. How many ionisable chlorine atoms are present in the formula of the compound?

Moles of AgCl=143.51.435​=0.01;50mL of 0.2M complex =0.01 mol [Co(NH3​)4​Cl2​]Cl→[Co(NH3​)4​Cl2​]++Cl−

Moles of $A g Cl = \frac{1.435}{143.5} = 0.01; 50 m L$ of $0.2 M$ complex $= 0.01$ mol
$[C o (N H_{3})_{4} C l_{2}] Cl \to [C o (N H_{3})_{4} C l_{2}]^{+} + C l^{-}$