Question

The total number of ions present in 1 ml of 0.1m barium nitrate solution is

Answer 1

Answer:$3\times 10^{-4}M$

Explanation:

$\text{no of moles of barium nitrate}={\text{Molarity}\times {\text{Volume in L}}$

$\text{no of moles of barium nitrate}={0.1M}\times {0.001L}=10^{-4}moles$             (1L=1000ml)

$Ba(NO_3)_2\rightarrow Ba^{2+}+2NO_3^{-1}$

1 mole of $Ba(NO_3)_2$ gives 1 mole of $Ba^{2+}$ and 2 moles of $NO_3^{-1}$

Thus $10^{-4}$ moles of  $Ba(NO_3)_2$ gives $10^{-4}[\tex]moles of [tex]Ba^{2+}$ and $2\times 10^{-4$ moles of $NO_3^{-1}$

Thus total ions are =$10^{-4}+2\times 10^{-4}=3\times 10^{-4}M$