Question

A sample of sodium sulfite has a mass of 2.80 g.

a) How many Na^+ ions are present?
b) How many SO_3^2- ions are present?
c) What is the mass in grams of one formula unit of Na_2SO_3?

Answer 1

Answer:

For a: The number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b: The number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c: The mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

Explanation:

The chemical formula of sodium sulfite is  $Na_2SO_3$. It is formed by the combination of 2 sodium $(Na^+)$ ions and 1 sulfite $(SO_3^{2-})$ ions

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium sulfite}=\frac{2.80g}{126g/mol}=0.022mol$

• For a:

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(2\times 0.022\times 6.022\times 10^{23})=2.65\times 10^{22}$ number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

• For b:

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(1\times 0.022\times 6.022\times 10^{23})=1.325\times 10^{22}$ number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

• For c:

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

$6.022\times 10^{23}$ number of formula units are present in 1 mole of a compound

Or, $6.022\times 10^{23}$ number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of = $\frac{126}{6.022\times 10^{23}}\times 1=2.09\times 10^{-22}g$

Hence, the mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams