What is the molality of a solution containing 11.1 g Na2SO4 dissolved in 2269.2 g of water?
Just making sure would it be 11.1 divided by 2269.2 times 100?
No. See my response below.
m = mols solute/kg solvent.
To calculate molality, you need to divide the moles of solute by the mass of the solvent in kilograms.
To find the moles of Na2SO4, you need to first calculate its molar mass.
The molar mass of Na2SO4 can be calculated by adding the atomic masses of all the atoms present. Looking up the atomic masses of the elements:
- Sodium (Na) has a molar mass of 22.99 g/mol.
- Sulfur (S) has a molar mass of 32.07 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
For sodium sulfate (Na2SO4), we have:
2(Na) + 1(S) + 4(O) = 2(22.99 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 142.04 g/mol.
Now, we can calculate the moles of Na2SO4:
moles = grams/molar mass = 11.1 g / 142.04 g/mol ≈ 0.0781 mol.
Next, convert the mass of water from grams to kilograms:
mass of water = 2269.2 g / 1000 = 2.2692 kg.
Finally, calculate the molality by dividing the moles of solute by the mass of the solvent in kilograms:
molality = moles / mass of water (in kg) = 0.0781 mol / 2.2692 kg ≈ 0.0344 mol/kg.
So, the molality of the solution is approximately 0.0344 mol/kg.
To summarize the calculation:
1. Calculate the molar mass of Na2SO4.
2. Determine the moles of Na2SO4.
3. Convert the mass of water from grams to kilograms.
4. Divide the moles of Na2SO4 by the mass of water in kilograms to find the molality.