Question

How many SO32- ions are contained in 99.6 mg of Na2SO3? The molar mass of Na2SO3 is 126.05 g/mol.

Answer 1

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, $47.57\times 10^{19}$

Answer 2

99.6 mg= .0996 g
.0996g x (1mol/125.05g) x (1 mol SO3/1 mol Na2SO2) x (6.02 x 10^22/1mol SO3) = 4.79 x 10^19 SO32- ions