Question

Solution containing 0.1 mole of a metal chloride ${MCl}_x$ requires 500 mL of 0.8 M ${AgNO}_3$ solution for complete precipitation. The value of $x$ is:

• A. 1
• B. 2
• C. 4
• D. 3

Answer 1

Answer: (C)

4

For complete precipitation, equal no. of moles should react by using formula,

$M_1{V}_1\times n{f}_1=M_2{V}_2\times n{f}_2\left(Mg{Cl}_x\right)\left(AgN{O}_3\right)$

For $MgC{l}_x$, $nf=x$

For $AgN{O}_3$, $nf=x$

$M\times V=$ no. of moles

Thus,

$0.1mole\times x=0.8M\times \frac{500L}{1000}\times 1$

$x=4$