Question

How many cations are there in 40.0 g of sodium phosphate?

Answer 1

Formula of sodium phosphate is $Na_3PO_4$

The molar mass of compound,$M = 164 g /mol$

Mass of compound, m = 30 g

Number of mol is mass of compound divided by its molar mass as below:

n = m /M

$= 30 / 164 = 0.182 mol$

Mole can be expressed as:

Number of particle /N_A = n = 0.182 mol

Number of particle $= 0.182 * 6.02 * 10^23$

$= 1.1 * 10^23 molecule$

As 3 sodium or cation are present in compound$= 3 * 1.1 * 10^23 cation = 3.3 * 10^23 cation$

Thus, 3.3 * 10^23  cation are present in 30 g sodium phosphate

Answer 2

There are 4.41 × 10²³ cations in 40.0 g of sodium phosphate.

First, we will convert 40.0 g of Na₃PO₄ to moles using its molar mass (163.94 g/mol).

$40.0 g \times \frac{1 mol}{163.94 g} = 0.244 mol$

Then, we will convert 0.244 moles to molecules using Avogadro's number: there are 6.02 × 10²³ molecules of Na₃PO₄ in 1 mole of molecules of Na₃PO₄.

$0.244 mol \times \frac{6.02 \times 10^{23} molecule}{mol} = 1.47 \times 10^{23} molecule$

Sodium phosphate is a strong electrolyte that dissociates according to the following equation.

Na₃PO₄ ⇒ 3 Na⁺ + PO₄³⁻

As we can see, there are 3 Na⁺ cations per molecule of Na₃PO₄. The number of  Na⁺ cations in 1.47 × 10²³ molecules of Na₃PO₄ are:

$1.47 \times 10^{23} molecule \times \frac{3Na^{+}cation }{1 molecule} = 4.41 \times 10^{23} Na^{+}cation$

There are 4.41 × 10²³ cations in 40.0 g of sodium phosphate.

You can learn more about Avogadro's number here:

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