CALCULATE THE MOLALITY OF THE AQUEOUS SOLUTION:

Question

CALCULATE THE MOLALITY OF THE AQUEOUS SOLUTION:

1a) 48.2 percent by mass KBr solution

1b) 1.22 M sugar (C₁₂H₂₂O₁₁) density = 1.12 g/mL

1c) 1. 87 M NaOH density = 1.04 g/mL

Number 2:

Determine the percent composition by mass of LiCl in a solution prepared by dissolving 2.34 g of LiCl in 67.5 g of water.

 

Answer 1

Molality (m) = moles of solute per kg of solvent (not to be confused with molarity)

1a) 4.2% (m/m) = 4.2 g KBr/100 g solution. We need moles KBr and kg of water, so we proceed as follows;

moles KBr = 4.2 g KBr x 1 mol KBr/119 g = 0.0353 moles KBr

kg water = 100 g solution - 4.2 g KBr = 95.8 g H2O = 0.0958 kg H2O

molality KBr = 0.0353 moles / 0.0958 kg = 0.368 m

1b) 1.22 M C₁₂H₂₂O₁₁ (sugar) , density 1.12 g/ml

Assume you have 1 liter of solution. Thus, you have 1.22 moles sugar. Now we need to find kg of H₂O.

moles sugar = 1.22 moles

mass of sugar = 1.22 moles x 342 g /mol = 417 g sugar

total mass of solution = 1 L x 1000 ml/L x 1.12 g/ml = 1120 g

mass of H₂O = 1120 g - 417 g = 703 g H₂O = 0.703 kg H₂O

molality of sugar = 1.22 mol / 0.703 kg = 1.74 m

1c) 1.87 M NaOH; density 1.04 g/ml

Assume you have 1 liter of solution. Thus, you have 1.87 moles NaOH. Now, find kg of H₂O.

moles NaOH = 1.87

mass NaOH = 1.87 moles NaOH x 40.0 g/mol = 74.8 g

total mass of solution = 1 L x 1000 ml/L x 1.04 g/ml = 1040 g

mass of H₂O = 1040 g - 74.8 g = 965 g = 0.965 kg H₂O

molality of NaOH = 1.87 mol / 0.965 = 1.94 m

2). Percent composition by mass is the mass of each component divided by the total mass (x100%). So, we need to find the total mass and also the mass of each component.

Total mass = 2.34 g + 67.5 g = 69.84 g

mass of LiCl = 2.34 g

mass of H₂O = 67.5 g

% by mass LiCl = 2.34 g / 69.84 g (x100%) = 3.35%

% by mass H₂O = 67.5 g / 69.8 g (x100%) = 96.7%

Now, if the question meant to ask for the % composition of Li⁺, Cl^- and H₂O, then you would find mass Li, mass Cl and follow the procedure above.

LiCl molar mass = 42.39 g/mol

moles LiCl = 2.34 g x 1 mol/42.39 g = 0.0552 moles LiCl