16.4g of solute is present in 500mL of the solution.
Explanation:
Given:
Volume of solution, V = 500mL = 0.5L
Molarity of Sodium phosphate, M = 0.2M
Molecular weight of Na₃PO₄ = 164 g/mol
Mass of the solute, m = ?
We know:
[tex]Molarity = \frac{moles of the solute}{Volume of solution in litres}[/tex]
And
Number of moles = [tex]\frac{mass of the solute}{molecular weight}[/tex]
On substituting the value we get:
[tex]0.2 = \frac{m}{164} X \frac{1}{0.5} \\\\m = 0.2 X 164 X 0.5\\\\m = 16.4 g[/tex]
Therefore, 16.4g of solute is present in 500mL of the solution.
