Answer:
The molarity of the sulfuric acid in the solution is 1.77 M.
Explanation:
The molarity of the sulfuric acid in the solution can be found using the following equation:
[tex] C_{i}V_{i} = C_{f}V_{f} \rightarrow C_{f} = \frac{C_{i}V_{i}}{V_{f}} [/tex]
Where:
[tex]C_{i}[/tex]: is the initial concentration of the acid
[tex]V_{i}[/tex]: is the initial volume of the solution = 22.6 cm³
[tex]V_{f}[/tex]: is the final volume of the solution = 88.5 cm³
The initial concentration of the H₂SO₄ is:
[tex] C_{i} = \frac{n}{V} = \frac{m}{M*V} = \frac{d*\% ^{m}_{m}}{M} [/tex]
Where:
n: is the number of moles
m: is the mass
M: is the molar mass = 98.079 g/mol
d: is the density of the acid = 1.39 g/cm³
%: is the percent by mass = 49.0 %
[tex] C_{i} = \frac{1.39 \frac{g}{cm^{3}}*\frac{1000 cm^{3}}{1 L}*\frac{49 g}{100 g}}{98.079 \frac{g}{mol}} = 6.94 M [/tex]
Finally, the final concentration of H₂SO₄ after the dilution is:
[tex] C_{f} = \frac{6.94 M*22.6 cm^{3}}{88.5 cm^{3}} = 1.77 M [/tex]
Therefore, the molarity of the sulfuric acid in the solution is 1.77 M.
I hope it helps you!
