Answer:
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Explanation:
Mass percentage of oxalic acid = 5.50%
This means that in 100 grams of solution there are 5.50 grams of oxalic acid.
Mass of solution , m = 100
Volume of the solution = V
Density of the solution = d = 1.024 g/mL
[tex]V=\frac{m}{d}=\frac{100 g}{1.024 g/mL}=97.66mL[/tex]
V = 97.66 mL = 0.09766 L
(1 mL = 0.001 L)
Moles of oxalic acid = [tex]\frac{5.50 g}{90 g/mol}=0.06111 mol[/tex]
[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex]
The molarity of the solution :
[tex]=\frac{0.06111 mol}{0.09766 L}=0.6257M[/tex]
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
