<h3>
Answer:</h3>
0.523 M
<h3>
Explanation:</h3>
The reaction between barium chloride and sodium sulfate is given by
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
We are given,
Volume of Na₂SO₄ = 15.0 mL
Mass of of solid BaSO₄ = 1.83 g
Required to determine the molarity of Na₂SO₄ solution
we will use the following steps
<h3>Step 1: Determine moles of the solid BaSO₄</h3>
Mass of BaSO₄ = 1.83 g
To get the number of moles we divide mass by the molar mass
Molar mass of BaSO₄ = 233.38 g/mol
Number of moles = 1.83 g ÷ 233.38 g/mol
= 0.00784 moles
<h3>Step 2: Moles of sodium sulfate used </h3>
From the balanced equation for every 1 mole of sodium sulfate used 1 mole of BaSO₄ was produced.
Therefore, the mole ratio of Na₂SO₄ : BaSO₄ is 1 : 1
Hence, moles of Na₂SO₄ will also be 0.00784 moles
<h3>Step 3: Molarity of sodium sulfate solution </h3>
Volume of sodium sulfate = 15.0 mL or 0.015 L
Number of moles = 0.00784 moles
But, molarity = Moles ÷ Volume
= 0.00784 moles ÷ 0.015 L
= 0.5227 M
= 0.523 M
Thus, the molarity of original sodium sulfate is 0.523 M
