Answer : The number of \(SO_3^{2-}\) ions are, \(47.57\times 10^{19}\)
Explanation : Given,
Mass of \(Na_2SO_3\) = 99.6 mg = 0.0996 g
Conversion used : (1 mg = 0.001 g)
Molar mass of \(Na_2SO_3\) = 126.05 g/mole
First we have to calculate the moles of \(Na_2SO_3\).
\(\text{Moles of }Na_2SO_3=\frac{\text{Mass of }Na_2SO_3}{\text{Molar mass of }Na_2SO_3}=\frac{0.0996g}{126.05g/mole}=7.90\times 10^{-4}moles\)
Now we have to calculate the number of \(SO_3^{2-}\).
As we know that, \(Na_2SO_3\) dissociate to give 2 moles of \(Na^+\) and 1 mole of \(SO_3^{2-}\).
From this we conclude that,
As, 1 mole contains \(6.022\times 10^{23}\) number of \(SO_3^{2-}\) ions
So, \(7.90\times 10^{-4}moles\) mole contains \((7.90\times 10^{-4})\times (6.022\times 10^{23})=47.57\times 10^{19}\) number of \(SO_3^{2-}\) ions
Therefore, the number of \(SO_3^{2-}\) ions are, \(47.57\times 10^{19}\)
