The volume (in mL) of 0.1 M AgNO_3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H_2O)_5Cl]Cl_2, as silver chloride is close to.

Question

Accepted Answer

Correct answer: 6. 6 Sol. m moles of [Cr(H_2O)_5Cl]Cl_2 = 0.01 × 30 = 0.3. ⇒ mmole of Cl^– = 0.3 × 2 = 0.6 ⇒ mmole of Ag^+ = mmoles of Cl^– ⇒ 0.1 × V = 0.6 ⇒ V = 6 mL.

$The volume (in mL) of 0.1 M A g N O_{3} required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr (H_{2} O)_{5} Cl] C l_{2}, as silver chloride is close to.$

Solution

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