At 25C an aqueous solution containing 300 wt H2SO4 has a spe

At 25°C, an aqueous solution containing 30.0 wt% H₂SO₄ has a specific gravity of 1.2150. A quantity of the 30.0% solution is needed that contains 345.5 kg of H₂SO₄.

(a). Calculate the required volume (L) of the solution using the given specific gravity.

(b). Estimate the percentage error that would have resulted if pure-component specific gravities of H₂SO₄ (1.8255 at 25°C) and water had been used for the calculation instead of the given specific gravity of the mixture.

(c). What is the specific gravity of a 30.0 wt% H₂SO₄ solution if there is no change in volume upon mixing?

Solution

30.0 wt% means this.. 100.0g solution contains 30.0g H2SO4..

And specific gravity means this. density of solution / (density of H2O @ THE SAME TEMPERATURE)
meaning this.... 1.2150 = (??? g/cm³) / (0.99704g/cm³)... where 0.99704 g/cm³ is the density of water at 25°C.

Then density of solution = 1.2150 x 0.99704 g/cm³ = 1.2114 g / cm³...
make sense?

Ans: a) so what you really need for volume is this...
345.5 kg H2SO4 x (1000g / 1kg) x (100.0g solution / 30.0g H2SO4) x (1 cm³ / 1.2150 g solution) = 9.478 x10^5 cm³ = 9.478x10^5 mL = 947.8 L

Ans: b).

start with that equation from above...
345.5 kg H2SO4 x (1000g / 1kg) x (100.0g solution / 30.0g H2SO4) x (1 cm³ / 1.2150 g solution)
......... .............. ......... .......... .......... ........ ........ ....... .......... ......... .......
....... ......... ........... .......... ............. ..and substitute 1 of 3 terms for density here..

density solution = 1.8255 x 0.99704 g/cm³ = 1.8201 g / cm³
density solution = 1.0000 x 0.99704 g/cm³ = 0.99704 g / cm³

Then volume of H2SO4 are.

345.5 kg H2SO4 x (1000g / 1kg) x (100.0g solution / 30.0g H2SO4) x (1 cm³ / 1.8201 g solution) = 632.7 L

345.5 kg H2SO4 x (1000g / 1kg) x (100.0g solution / 30.0g H2SO4) x (1 cm³ / 0.99704 g solution) = 1155.08 L

and % errors would be

| 632.7 - 947.8 | / | 947.8 | x 100% = 33.24%
| 1155.08 - 947.8 | / | 947.8 | x 100% = 21.86%

Ans: c)

100g solution x (30.0 g H2SO4 / 100g solution) x (1 cm³ H2SO4 / 1.8201 g H2SO4) = 16.48 cm³ H2SO4
100g solution x (70.0 g H2O / 100g solution) x (1 cm³ H2O / 0.99704 g H2O) = 70.207 cm³ H2O
total volume = 86.687 mL.. density = 100.000g / 86.687 mL = 1.1535 g / mL... then specific gravity 1.157 (Specific gravity is density of solution to density of water at same temperature).