Answer :
The correct answer is for mass of Na₃PO₄ 39.7 g.
Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00
2) Volume of solution = 725 mL
Converting volume of solution from mL to L :
Conversion factor : 1 L = 1000 mL
Volume of solution = 0.725 L
Following steps can be done to find mass of :
<u>Step 1 : Write the dissociation reaction of Na₃PO₄ .</u>
<u>Step 2: Find moles of Na⁺ ions : </u>
Mole of Na⁺ ions can be calculated using molarity formula which is :
Plugging value of Molarity and volume
Multiplying both side by 0.725 L
<em>Mole of Na⁺ ions = 0.725 mol</em>
<u>Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :</u>
Mole ratio is found from coefficients from balanced reaction as:
Mole of Na₃PO₄ in balanced reaction = 1
Mole of Na⁺ ion = 3
<em>Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3 </em>
<u>Step 4 : To find mole of Na₃PO₄ </u>
Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :
<em>Mole of Na₃PO₄ = 0.242 mol </em>
<u>Step 5 : To find mass of Na₃PO₄</u>
Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :
Mass of Na₃PO₄ = 39.619 g
<u>Step 6 : To round off mass of Na₃PO₄ to correct sig fig .</u>
The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :
Mass of Na₃PO₄ = 39.7 g