Question

If the molar concentration of ${PbI}_2$ is $1.5\times {10}^{-3}mol{L}^{-1}$, then the concentration of iodide ions in g ion $L^{-1}$ is:

• A. $3.0\times {10}^{-3}$
• B. $6.0\times {10}^{-3}$
• C. $0.3\times {10}^{-3}$
• D. $0.6\times {10}^{-6}$

Answer 1

The correct option is A $3.0\times {10}^{-3}$

${PbI}_2\to {Pb}^{2+}+2I^{(-)}$

Since 1 mole ${PbI}_2\to 2mole{I}^{(-)}$

$\therefore 1.5\times {10}^{-3}{molL}^{-1}\to 2\times 1.5\times {10}^{-3}mol{L}^{-1}$

$=3\times {10}^{-3}mol{L}^{-1}$