Calculate the volume of concentrated sulphuric acid of relative density 1.84 and containing 98% sulphuric acid by mass that would contain 40g of pure sulphuric acid.
Answers
Answer:
Explanation:
method-1
The density of any substance is given in g/cm^3. Not in g/mol as that would be the molecular weight. So the density is 1.84 g/cm^3 (=1.84 g/mL).
The answer:
0.2M = 0.2 mole/L
Therefore, 0.5L * 0.2M = 0.1 mole of H2SO4 required.
The molar mass of H2SO4 is 98.079 g/mole (see periodic table or google).
Therefore, 98.079 g/mole * 0.1 mole = 9.8079 g
9.8079 g/ 1.84 g/cm^3 = 5.33038 cm^3 (=5.33038 mL)
The purity of your substance is 96%, therefore you will actually need:
5.33038*(1/0.96) = 5.55248 mL.
The required precision in your answer is 3 digits (due to the precision of 500 mL).
Therefore, the answer is 5.55 mL of acid and (500–5.55 =) 494 mL of water.
Note that when preparing such a dilution, always add acid to water. Never the other way around. Pure H2SO4 is an extremely dangerous substance. Handle only under the supervision of a professional.
method-2
Molar mass of sulphuric acid = 98
500 ml of 0.2000 M sulphuric acid= 0.1000 moles = 4.9 g
Since the solution is 96% of sulphuric acid = hence 1 ml will contain 0.96*1.84g = 1 .7664 g
Hence volume of solution needed = 4.9 ÷ 1.7664 = 2.77 ml