Question

The number of chloride ions present in $0.5$ mole of $MgC{l}_2$ is:

• A. $2N_A$
• B. $N_A/2$
• C. $N_A/4$
• D. $N_A$

Answer 1

The correct option is D $N_A$

$MgC{l}_2\to M{g}^{2+}+2C{l}^{-}$

Number of moles of chloride ions present in 1 mole of $MgC{l}_2=2$

Therefore, number of moles of chloride ions present in 0.5 moles of $MgC{l}_2$,

Number of chloride ions$=0.5\times N_A\times 2$ chloride ions.

Hence, option D is correct.