To determine the molarity of a [tex]$20.0 \%$[/tex] by mass solution of phosphoric acid ([tex]$H_3PO_4$[/tex]) in water with a density of 1.114 g/mL at [tex]$20^{\circ} C$[/tex], follow these steps:
1. Determine the mass of the solution:
- Given that the density of the solution is 1.114 g/mL, we first convert 1 liter of solution to milliliters, which is 1000 mL.
- The mass of 1 liter (1000 mL) of the solution is then:
[tex]\[
\text{Mass of solution} = \text{Density} \times \text{Volume} = 1.114 \text{ g/mL} \times 1000 \text{ mL} = 1114 \text{ g}
\][/tex]
2. Calculate the mass of [tex]$H_3PO_4$[/tex] in the solution:
- The solution is 20.0% by mass [tex]$H_3PO_4$[/tex], which means 20.0% of the total mass of the solution is [tex]$H_3PO_4$[/tex].
- Therefore, the mass of [tex]$H_3PO_4$[/tex] in 1114 g of solution is:
[tex]\[
\text{Mass of } H_3PO_4 = \text{Mass percent} \times \text{Mass of solution} = 0.20 \times 1114 \text{ g} = 222.8 \text{ g}
\][/tex]
3. Calculate the moles of [tex]$H_3PO_4$[/tex]:
- The molar mass of [tex]$H_3PO_4$[/tex] is given as 97.994 g/mol.
- To find the number of moles of [tex]$H_3PO_4$[/tex], use the formula:
[tex]\[
\text{Moles of } H_3PO_4 = \frac{\text{Mass of } H_3PO_4}{\text{Molar mass of } H_3PO_4} = \frac{222.8 \text{ g}}{97.994 \text{ g/mol}} \approx 2.274 \text{ mol}
\][/tex]
4. Calculate the molarity of the solution:
- Molarity (M) is defined as the number of moles of solute per liter of solution.
- Since we are considering 1 liter of solution, the molarity is simply the number of moles of [tex]$H_3PO_4$[/tex] calculated:
[tex]\[
\text{Molarity of } H_3PO_4 = \frac{\text{Moles of } H_3PO_4}{\text{Volume of solution in liters}} = \frac{2.274 \text{ mol}}{1 \text{ L}} = 2.274 \text{ M}
\][/tex]
Thus, the molarity of the phosphoric acid ([tex]$H_3PO_4$[/tex]) solution is approximately 2.27 M. The correct answer is:
D) 2.27 M
