Formula of sodium phosphate is [tex]Na_3PO_4[/tex]
The molar mass of compound,[tex]M = 164 g /mol[/tex]
Mass of compound, m = 30 g
Number of mol is mass of compound divided by its molar mass as below:
n = m /M
[tex]= 30 / 164 = 0.182 mol[/tex]
Mole can be expressed as:
Number of particle /N_A = n = 0.182 mol
Number of particle [tex]= 0.182 * 6.02 * 10^23[/tex]
[tex]= 1.1 * 10^23 molecule[/tex]
As 3 sodium or cation are present in compound[tex]= 3 * 1.1 * 10^23 cation = 3.3 * 10^23 cation[/tex]
Thus, 3.3 * 10^23 cation are present in 30 g sodium phosphate
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).
[tex]40.0 g \times \frac{1 mol}{163.94 g} = 0.244 mol[/tex]
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.
[tex]0.244 mol \times \frac{6.02 \times 10^{23} molecule}{mol} = 1.47 \times 10^{23} molecule[/tex]
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:
[tex]1.47 \times 10^{23} molecule \times \frac{3Na^{+}cation }{1 molecule} = 4.41 \times 10^{23} Na^{+}cation[/tex]
There are 4.41 × 10²³ cations in 40.0 g of sodium phosphate.
You can learn more about Avogadro's number here:
https://brainly.com/question/13302703
