Answer :
The correct answer is for mass of Na₃PO₄ 39.7 g.
Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 ![\frac{mol}{L}](https://tex.z-dn.net/?f=%5Cfrac%7Bmol%7D%7BL%7D%20)
2) Volume of solution = 725 mL
Converting volume of solution from mL to L :
Conversion factor : 1 L = 1000 mL
![Volume of solution = 725 mL * \frac{1L }{1000 mL}](https://tex.z-dn.net/?f=Volume%20of%20solution%20%3D%20725%20mL%20%2A%20%5Cfrac%7B1L%20%7D%7B1000%20mL%7D)
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>
![Na_3PO_4 3 Na^+ + PO_4^3^-](https://tex.z-dn.net/?f=Na_3PO_4%20%20%3C%3D%3E%20%203%20Na%5E%2B%20%20%2B%20%20PO_4%5E3%5E-)
<u>Step 2: Find moles of Na⁺ ions : </u>
Mole of Na⁺ ions can be calculated using molarity formula which is :![Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }](https://tex.z-dn.net/?f=Molarity%20%28%5Cfrac%7Bmol%7D%7BL%7D%20%29%20%3D%20%5Cfrac%7Bmole%20of%20Na%5E%2B%20%7D%7Bvolume%20of%20solution%20%7D)
Plugging value of Molarity and volume
![1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}](https://tex.z-dn.net/?f=1.00%20%5Cfrac%7Bmol%7D%7BL%7D%20%3D%20%5Cfrac%7BMole%20of%20Na%5E%2B%20ions%7D%7B0.725%20L%7D)
Multiplying both side by 0.725 L
![1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L](https://tex.z-dn.net/?f=1.00%20%5Cfrac%7Bmol%7D%7BL%7D%2A%200.725%20L%20%3D%20%5Cfrac%7Bmole%20of%20Na%5E%2B%20ions%7D%7B0.725%20L%7D%20%2A%200.725%20L)
<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 :
![Mole of Na_3PO_4= Mole of Na^+ * Mole ratio](https://tex.z-dn.net/?f=Mole%20of%20Na_3PO_4%3D%20Mole%20of%20Na%5E%2B%20%20%2A%20Mole%20ratio)
![Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }](https://tex.z-dn.net/?f=Mole%20of%20Na_3PO_4%20%3D%200.725%20mol%20%20%2A%20%20%5Cfrac%7B1%20mole%20of%20Na_3PO_4%7D%7B3%20mole%20of%20Na%5E%2B%20%7D)
<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 (g) = mole (mol) * molar mass \frac{g}{mol}](https://tex.z-dn.net/?f=Mass%20%28g%29%20%3D%20mole%20%28mol%29%20%2A%20molar%20mass%20%5Cfrac%7Bg%7D%7Bmol%7D)
![Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}](https://tex.z-dn.net/?f=Mass%20of%20Na_3PO_4%20%3D%20%200.242%20mol%20%2A%20163.94%20%5Cfrac%7Bg%7D%7Bmol%7D)
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