Respuesta :
Answer :
The correct answer is for mass of Na₃PO₄ 39.7 g.
Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 [tex]\frac{mol}{L} [/tex]
2) Volume of solution = 725 mL
Converting volume of solution from mL to L :
Conversion factor : 1 L = 1000 mL
[tex]Volume of solution = 725 mL * \frac{1L }{1000 mL}[/tex]
Volume of solution = 0.725 L
Following steps can be done to find mass of :
Step 1 : Write the dissociation reaction of Na₃PO₄ .
[tex]Na_3PO_4 <=> 3 Na^+ + PO_4^3^-[/tex]
Step 2: Find moles of Na⁺ ions :
Mole of Na⁺ ions can be calculated using molarity formula which is :[tex]Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }[/tex]
Plugging value of Molarity and volume
[tex]1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}[/tex]
Multiplying both side by 0.725 L
[tex]1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L[/tex]
Mole of Na⁺ ions = 0.725 mol
Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :
Mole ratio is found from coefficients from balanced reaction as:
Mole of Na₃PO₄ in balanced reaction = 1
Mole of Na⁺ ion = 3
Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3
Step 4 : To find mole of Na₃PO₄
Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :
[tex]Mole of Na_3PO_4= Mole of Na^+ * Mole ratio[/tex]
[tex]Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }[/tex]
Mole of Na₃PO₄ = 0.242 mol
Step 5 : To find mass of Na₃PO₄
Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :
[tex]Mass (g) = mole (mol) * molar mass \frac{g}{mol}[/tex]
[tex]Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}[/tex]
Mass of Na₃PO₄ = 39.619 g
Step 6 : To round off mass of Na₃PO₄ to correct sig fig .
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
Tribasic Sodium phosphate is widely used as a cleaning agent and helps to prevent the growth of bacteria. The grams of tribasic sodium phosphate required by the scientist is 39.619 or 39.7 grams.
Given that,
- Molarity of sodium ions = 1.00 M
- Volume of the solution = 725 mL or 0.725 L
- Now, calculating the moles of ions from the balanced equation:
- Na[tex]_3[/tex]PO[tex]_4[/tex] ⇔ 3 N[tex]\text a^+[/tex] + PO[tex]_4^{3-}[/tex].
- [tex]\begin{aligned}\text {Molarity}&=\dfrac{\text{moles of Na}^+}{\text{Volume of solution}}\\\\\text {1.00}&=\dfrac{\text{moles of Na}^+}{\text{0.725}}\\\\\text {1.00} \times 0.725}&=\dfrac{\text{moles of Na}^+}{\text{0.725}} \times 0.725 \end[/tex]
- Number of moles of sodium ions = 0.725 moles
Now,
The mole ratio of tribasic sodium phosphate can be calculated as :
- Mass of Na[tex]_3[/tex]PO[tex]_4[/tex] = Moles of N[tex]\text a^+[/tex] x Mole ratio
- Mass of Na[tex]_3[/tex]PO[tex]_4[/tex] = 0.725 x [tex]\dfrac {\text{1 mole of Na}_3 \text{PO}_4}{\text {3 moles of Na}^+}[/tex]
- Mass of Na[tex]_3[/tex]PO[tex]_4[/tex] = 39.619 gram.
Now, rounding off the mass of tribasic sodium phosphate, it can be written as:
Mass of Na[tex]_3[/tex]PO[tex]_4[/tex] = 39. 7 g
Therefore, the mass of Na[tex]_3[/tex]PO[tex]_4[/tex] required by the scientist is approximately 39.7 g.
To know more about molarity, refer to the following link:
https://brainly.com/question/19567514?referrer=searchResults
