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

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:

$C_{i}V_{i} = C_{f}V_{f} \rightarrow C_{f} = \frac{C_{i}V_{i}}{V_{f}}$    

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:

$C_{i} = \frac{n}{V} = \frac{m}{M*V} = \frac{d*\% ^{m}_{m}}{M}$

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 %

$C_{i} = \frac{1.39 \frac{g}{cm^{3}}*\frac{1000 cm^{3}}{1 L}*\frac{49 g}{100 g}}{98.079 \frac{g}{mol}} = 6.94 M$          

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

$C_{f} = \frac{6.94 M*22.6 cm^{3}}{88.5 cm^{3}} = 1.77 M$

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

I hope it helps you!