Question

A solution containing 0.1 mol of a metal chloride $MCl_x$ requires 500 mol. of 0.8 M $AgNO_3$ solution for complete precipitation $MCl_x+xAgNO_3 \to xAgCl+M(NO_3)_x$. The value of x is

• A. 1
• B. 2
• C. 4
• D. 3

Answer 1

Answer: (C)

4

Answer 2

$Mclx+xAgNO_{3} \to xAgCl+M(NO)_{3}x$ Moles of $AgNO_{3}$ required for precipitation $=500\times0.8\times10^{-3}$ Now $0.1$ mol of $MCl_{x}$ requires $AgNO_{3} = 0.4\, mol$ $1$ mol of $MCl_{x}$ requires $AgNO_{3} = 4 \,mol$ Hence $x = 4$