Molality of sulphuric acid is 8.01m and its density is 1.354g/ml. Find molarity and mole fraction of sulphuric acid?
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 = ?
To find the molarity and mole fraction of sulphuric acid, we can use the given information about the molality and density.
First, we need to convert the density from grams per milliliter (g/ml) to grams per cubic centimeter (g/cm³) since molality is defined as the number of moles of solute per kilogram of solvent.
Density = 1.354 g/ml
We know that 1 ml = 1 cm³, so the density in g/cm³ would be:
1.354 g/ml = 1.354 g/cm³
Now, let's calculate the molarity.
Molarity (M) is defined as the number of moles of solute per liter of solution. To find the moles of solute, we need to know the molecular weight of sulphuric acid, which is H₂SO₄.
Molecular weight of H₂SO₄ = 2(1.008 g/mol) + 32.06 g/mol + 4(16.00 g/mol)
= 98.09 g/mol
The molality (m) is defined as the number of moles of solute per kilogram of solvent. We know the molality:
Molality (m) = 8.01 m
Since the density of the solution in g/cm³ is numerically equal to the mass of the solution in grams per cubic centimeter, we can say that the density of the solution in g/cm³ is equal to the mass of the solution (grams) divided by the volume of the solution (cm³).
Now, let's calculate the mass of the solution.
Given:
Density of the solution = 1.354 g/cm³
Volume of the solution = 1 L (since molarity is defined per liter of solution)
Mass of the solution = Density × Volume
= 1.354 g/cm³ × 1000 cm³ (since 1 L = 1000 cm³)
= 1354 g
Now that we know the mass of the solution, we can calculate the moles of solute.
Moles of solute = Molality × Mass of solvent (kg)
= 8.01 m × 1.354 kg (since 1 L = 1 kg for water, assuming it's the solvent)
= 10.85 mol
Now, we can find the molarity by dividing the moles of solute by the volume of the solution in liters.
Molarity (M) = Moles of solute / Volume of solution (L)
= 10.85 mol / 1 L
= 10.85 M
Finally, let's calculate the mole fraction.
Mole fraction is the ratio of the moles of solute to the total moles of solute and solvent in the solution.
Mole fraction of sulphuric acid = Moles of solute / (Moles of solute + Moles of solvent)
Since the molality (m) is defined as the number of moles of solute per kilogram of solvent, we can assume that the mass of the solvent is equal to the mass of the solution in this case.
Moles of solvent = Mass of solvent / Molecular weight of solvent
Mass of solvent (H₂O) = Mass of solution - Mass of solute
= 1354 g - (10.85 mol × 98.09 g/mol)
= 1354 g - 1066.86 g
= 287.14 g
Moles of solvent (H₂O) = 287.14 g / 18.015 g/mol (molecular weight of H₂O)
= 15.94 mol
Mole fraction of sulphuric acid = 10.85 mol / (10.85 mol + 15.94 mol)
= 10.85 mol / 26.79 mol
= 0.405
So, the molarity of sulphuric acid is 10.85 M and the mole fraction is 0.405.
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