Asked by Snehal bhoir
Molality of sulphuric acid is 8.01m and its density is 1.354g/ml. Find molarity and mole fraction of sulphuric acid?
Answers
Answered by
Arora
Molality = number of moles / mass of solvent
Consider 100ml of the given solution.
It has 135.4 g of sulphuric acid.
M.M. of sulphuric acid = 98g
Number of moles in this case = 135.4/98 = 1.31
Molarity = number of moles/volume of solution (in Liters)
= 1.31/0.1 = 13.1
Hence, the solution is 13.1M
Consider 100ml of the given solution.
It has 135.4 g of sulphuric acid.
M.M. of sulphuric acid = 98g
Number of moles in this case = 135.4/98 = 1.31
Molarity = number of moles/volume of solution (in Liters)
= 1.31/0.1 = 13.1
Hence, the solution is 13.1M
Answered by
DrBob222
Your solution is not correct. If the solution has a density of 1.354 g/mL then the mass of the solution, correctly, is 135.4g; however, that is the mass of the H2SO4 + the mass of the solvent. You are using it as the mass of the solvent only.
Instead of 100 mL I will assume enough solvent to contain 8.01 mols H2SO4. That will be 1000 mL SOLVENT(H2O). 8.01 mols x 98 = 785 g H2SO4 So the SOLUTION is 1000 g H2O + 785 g H2SO4 = 1785 grams of solution.
Then volume = mass/density = 1785/1.354 = about 1318 mL or 1.318 L.
Then M = mols/L = 8.01 mols/1.318 L = ?
Instead of 100 mL I will assume enough solvent to contain 8.01 mols H2SO4. That will be 1000 mL SOLVENT(H2O). 8.01 mols x 98 = 785 g H2SO4 So the SOLUTION is 1000 g H2O + 785 g H2SO4 = 1785 grams of solution.
Then volume = mass/density = 1785/1.354 = about 1318 mL or 1.318 L.
Then M = mols/L = 8.01 mols/1.318 L = ?
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