Question

A solution of 49.0% H2SO4 by mass has a density of 1.39 g cm−3 at 293 K. A 22.6 cm3 sample of this solution is mixed with enough water to increase the volume of the solution to 88.5 cm3 .

Find the molarity of sulfuric acid in this solution.

Answer 1

The molarity of the sulfuric acid in the solution is 1.77 M.

Step-by-step explanation:

The molarity of the sulfuric acid in the solution can be found using the following equation:

`C_(i)V_(i) = C_(f)V_(f) \rightarrow C_(f) = (C_(i)V_(i))/(V_(f))`

Where:

`C_(i)`: is the initial concentration of the acid

`V_(i)`: is the initial volume of the solution = 22.6 cm³

`V_(f)`: is the final volume of the solution = 88.5 cm³

The initial concentration of the H₂SO₄ is:

`C_(i) = (n)/(V) = (m)/(M*V) = (d*\% ^(m)_(m))/(M)`

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 %

`C_(i) = (1.39 (g)/(cm^(3))*(1000 cm^(3))/(1 L)*(49 g)/(100 g))/(98.079 (g)/(mol)) = 6.94 M`

Finally, the final concentration of H₂SO₄ after the dilution is:

`C_(f) = (6.94 M*22.6 cm^(3))/(88.5 cm^(3)) = 1.77 M`

Therefore, the molarity of the sulfuric acid in the solution is 1.77 M.

I hope it helps you!