If a student added 42.6 grams of sodium sulfite to 780.4 grams of water and determined that 3.40 grams of the new solution occupied 3.20ml, what is the weight percent, molality, and molarity of the solution produced?
And how much of this can you do? I can help you through this but I don't want to replicate what you already know how to do.
I figured it out. I finally realized that the density was given in the problem so I could convert from grams to ml to L for the molarity! Thank you for the offer to help me.
To determine the weight percent, molality, and molarity of the solution, we need to understand the definitions of these terms and perform some calculations.
1. Weight Percent:
Weight percent is the weight of the solute divided by the total weight of the solution, multiplied by 100.
To calculate the weight percent, we need to find the weight of the solute (sodium sulfite in this case) and the total weight of the solution.
Weight of solute = 42.6 grams
Weight of solution = Weight of solute + Weight of solvent
Weight of solvent (water) = 780.4 grams
Weight percent = (Weight of solute / Weight of solution) * 100
Weight percent = (42.6 / (42.6 + 780.4)) * 100
Weight percent = (42.6 / 823) * 100
Weight percent ≈ 5.17%
Therefore, the weight percent of the solution produced is approximately 5.17%.
2. Molality:
Molality is the number of moles of solute dissolved in 1 kilogram of solvent.
To calculate the molality, we need to calculate the number of moles of solute and the mass of the solvent.
Molar mass of sodium sulfite (Na2SO3) = 2 * atomic mass of Na + atomic mass of S + 3 * atomic mass of O
= 2 * 22.99 + 32.07 + 3 * 16.00
= 46.00 + 32.07 + 48.00
= 126.07 g/mol
Number of moles of solute = (Weight of solute / Molar mass of solute)
Number of moles = 42.6 / 126.07
Mass of solvent = Weight of solution - Weight of solute
Mass of solvent = 780.4 - 42.6
Molality = (Number of moles of solute / Mass of solvent in kg)
Molality = (42.6 / 126.07) / (737.8 / 1000)
Molality ≈ 0.0685 mol/kg
Therefore, the molality of the solution produced is approximately 0.0685 mol/kg.
3. Molarity:
Molarity is the number of moles of solute dissolved in 1 liter (1000 mL) of solution.
To calculate the molarity, we need to calculate the number of moles of solute and the volume of the solution.
Number of moles of solute = (Weight of solute / Molar mass of solute)
Number of moles = 42.6 / 126.07
Volume of solution = Volume of 3.40 grams of the solution
Molarity = (Number of moles of solute / Volume of solution in liters)
Molarity = (42.6 / 126.07) / (3.20 / 1000)
Molarity ≈ 0.133 mol/L
Therefore, the molarity of the solution produced is approximately 0.133 mol/L.
In summary:
- Weight percent ≈ 5.17%
- Molality ≈ 0.0685 mol/kg
- Molarity ≈ 0.133 mol/L