An aqueous solution that is 10.0 percent sulfuric acid (H2SO4) by mass has a density of 1.143 g/mL. Determine the molality of the solution. ( 3 significant figures)
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To determine the molality of the solution, we need to find the number of moles of sulfuric acid and the mass of the solvent.
First, let's find the mass of the sulfuric acid in the solution:
Given that the solution is 10.0% sulfuric acid by mass, we can express this as a mass fraction. This means that 10.0% of the mass of the solution is due to sulfuric acid.
Let's assume we have 100 g of the solution. Therefore, the mass of sulfuric acid in the solution is 10.0 g (10.0% of 100 g).
Next, we need to find the number of moles of sulfuric acid. To do this, we divide the mass of sulfuric acid by its molar mass.
The molar mass of H2SO4 can be calculated using the atomic masses of hydrogen (H) and sulfur (S), as well as the molar mass of oxygen (O), which can be found on the periodic table:
H: 1.00784 g/mol
S: 32.06 g/mol
O: 15.999 g/mol
Molar mass of H2SO4:
2(1.00784 g/mol) + 32.06 g/mol + 4(15.999 g/mol) = 98.087 g/mol
Now we can calculate the number of moles of sulfuric acid:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
moles of H2SO4 = 10.0 g / 98.087 g/mol
Once we know the number of moles of sulfuric acid, we need to find the mass of the solvent in the solution, which is the mass difference between the solution and the mass of sulfuric acid.
Given that the density of the solution is 1.143 g/mL, we can calculate the mass of the solution using the equation:
mass of solution = volume of solution * density of solution
Since we don't know the volume of the solution, we'll assume we have 1 L (1000 mL) of the solution for convenience:
mass of solution = 1000 mL * 1.143 g/mL = 1143 g
Now, let's find the mass of the solvent:
mass of solvent = mass of solution - mass of sulfuric acid
mass of solvent = 1143 g - 10.0 g
Finally, we can calculate the molality of the solution:
molality = moles of solute / mass of solvent
molality = moles of H2SO4 / mass of solvent
Substituting the calculated values, we have:
molality = (10.0 g / 98.087 g/mol) / (1133 g - 10.0 g)
Performing the calculations gives:
molality ≈ 0.101 mol/kg
Therefore, the molality of the solution is approximately 0.101 mol/kg (to 3 significant figures).