What is the molality of each of the following solutions?
1)Dissolve 0.610 mol of citric acid, C6H8O7, in 1.35 kg of water.
2) Dissolve 0.200 mg of KBr in 5.50 mL of water.
3) Dissolve 4.05 g of aspirin, C9H8O4, in 145 g of dichloromethane, CH2Cl2.
For some reason I keep forgetting how to find molarity? I know it has to do with converting grams to mols but I'm not sure. If someone could explain it simply and give an example?
1.
m = mols/kg solvent
m = 0.610/1.35 = ?
2.
0.200 mg KBr = 0.0002 g.
mols = grams/molar mass = 0.002/119 = 1.68E-6
m = mols/kg solvent = 1.68E-6/0.0055 = ?
etc.
To find the molality of a solution, you need to know the number of moles of solute and the mass of the solvent in kilograms. Molality is defined as the number of moles of solute per kilogram of solvent.
Here's how you can find the molality step by step:
1) Determine the number of moles of the solute. This can be done by dividing the given mass of the solute by its molar mass. The molar mass is the sum of the atomic masses of all the atoms in the formula. You can find the atomic masses in the periodic table.
2) Convert the mass of the solvent to kilograms. This can be done by dividing the given mass of the solvent by 1000.
3) Once you have the number of moles of solute and the mass of the solvent in kilograms, divide the number of moles of solute by the mass of the solvent in kilograms to find the molality.
Now, let's apply this to the given examples:
1) Dissolve 0.610 mol of citric acid, C6H8O7, in 1.35 kg of water.
- First, find the molar mass of citric acid:
C: 12.01 g/mol x 6 = 72.06 g/mol
H: 1.01 g/mol x 8 = 8.08 g/mol
O: 16.00 g/mol x 7 = 112.00 g/mol
Total molar mass of citric acid = 72.06 g/mol + 8.08 g/mol + 112.00 g/mol = 192.14 g/mol
- Convert the mass of water to kilograms:
1.35 kg (already in kg)
- Calculate the molality:
molality = 0.610 mol / 1.35 kg = 0.452 mol/kg
2) Dissolve 0.200 mg of KBr in 5.50 mL of water.
- First, convert the mass of KBr to grams:
0.200 mg = 0.200 × 10^(-3) g
- Convert the volume of water to kilograms:
5.50 mL = 5.50 × 10^(-3) L = 5.50 × 10^(-6) kg
- Calculate the molality:
molality = 0.200 × 10^(-3) g / 5.50 × 10^(-6) kg = 36.36 mol/kg
3) Dissolve 4.05 g of aspirin, C9H8O4, in 145 g of dichloromethane, CH2Cl2.
- First, find the molar mass of aspirin:
C: 12.01 g/mol x 9 = 108.09 g/mol
H: 1.01 g/mol x 8 = 8.08 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total molar mass of aspirin = 108.09 g/mol + 8.08 g/mol + 64.00 g/mol = 180.17 g/mol
- Convert the mass of dichloromethane to kilograms:
145 g = 145/1000 kg
- Calculate the molality:
molality = 4.05 g / 145/1000 kg = 27.93 mol/kg
By following these steps, you can calculate the molality of any given solution.