Question

The total number of ions present in 1 mL of 0.1 M barium nitrate $Ba{\left({NO}_3\right)}_2$ solution is:

• A. $6.02\times {10}^{18}$
• B. $6.02\times {10}^{19}$
• C. $3.0\times 6.02\times {10}^{19}$
• D. $3.0\times 6.02\times {10}^{18}$

Answer 1

Answer: (C)

$3.0\times 6.02\times {10}^{19}$

Number of moles $=molarity\times volume$

$=0.1\times 0.001$

$x={10}^{-4}$ moles or $0.0001$ moles

$Ba(N{O}_3{)}_2⟶B{a}^{2+}+{{NO}_3}^{2-}$

The total number of ions = $3$ ions in one mole of $Ba(N{O}_3{)}_2$.

So, in ${10}^{-4}$ moles = $3\times 6.023\times {10}^{23}\times {10}^{-4}$

= $3.0\times 6.02\times {10}^{19}$

Hence, option $C$ is correct.