A 500 ml aqueous solution of na3po4 was prepared using 82g of the solute. what is the molarity

The molarity of the solution is 1 M

From the question,

We are to determine the molarity of the solution.

First, we will determine the number of moles of the solute (Na₃PO₄) present

Mass of Na₃PO₄ = 82 g

Using the formula

Molar mass of Na₃PO₄ = 163.94 g/mol

∴ Number of moles of Na₃PO₄ present =

Number of moles of Na₃PO₄ present = 0.500 moles

Now, for the molarity of the solution

Using the formula

Volume of the solution = 500 mL = 0.5 L

∴ Molarity of the solution =

Molarity of the solution = 1 M

Hence, the molarity of the solution is 1 M

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1 M

Explanation:

The molarity of a solution, M, is a measure of the concentration of that solution and it refers to the number of moles of solute (mol) per liter (L) of solution. The molarity (M) can be calculated using the formula:

M = number of moles (n) /volume (V)

In this question, a 500 ml aqueous solution of Na3PO4 was prepared using 82g of the solute.

Molar mass of Na3PO4 = 23(3) + 15 + 16(4)

= 69 + 31 + 64

= 164g/mol

Mole = mass/molar mass

mole = 82/164

mole = 0.5 mol

Volume in Litres (L) = 500 ml ÷ 1000 = 0.500L

Therefore, Molarity (M) = 0.5/0.500

Molarity = 1 M or 1 mol/L

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