Answer:
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
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
[tex]Number\ of\ moles = \frac{Mass }{Molar\ mass}[/tex]
Molar mass of Na₃PO₄ = 163.94 g/mol
∴ Number of moles of Na₃PO₄ present = [tex]\frac{82}{163.94}[/tex]
Number of moles of Na₃PO₄ present = 0.500 moles
Now, for the molarity of the solution
Using the formula
[tex]Molarity = \frac{Number\ of\ moles}{Volume}[/tex]
Volume of the solution = 500 mL = 0.5 L
∴ Molarity of the solution = [tex]\frac{0.500}{0.5}[/tex]
Molarity of the solution = 1 M
Hence, the molarity of the solution is 1 M
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