Question

1.What is the molality of a solution consisting of 1.34 mL of CCl₄ (D=1.59 g/mL) in 65.0 mL of CH₂Cl₂ (D=1.33 g/mL).

2. Calculate the molarity of 3.58 m aqueous RbCl solution with a density of 1.12 g/mL.

3.An acidic solution containing 571.4g H₂SO₄ per liter has a density of 1.329 g/mL. Calculate the molality of H₂SO₄ in this solution.

Answer 1

1. Molarity = moles of solute/ Volume (L)

Amount of solute = 1.34 mL x 1.59 g/mL = 2.13 g

Molar mass of CCl₄ = 12 + 4(35.5) = 154 g/mol

moles of solute = 2.13 g /154 gmol ^-1 = 0.0138 mol

mass of solvent = 65.0 mL x 1.33g/mL = 86.45 g

m = moles/ kg of solvent = 0.0138 mol/0.08645 = 0.159m

molality = 0.159 m

2. 3.58 molal soln. of RbCl means 3.58 moles of RbCl are dissolved in 1 kg solvent. So the mass of solution

= 1000 g of solvent + 432.89 g RbCl = 1432.89 g of solution (molecular mass of RbCl: 432.89 g/mol)

Volume of solution = Mass of solution / Density of solution = 1432.89 g / 1.12 gmL^-1 = 1279.36 mL

Molarity = moles of solute x 1000/ V(mL) = 3.58 x 1000/1279.36 = 2.79 M

Molarity = 2.79 M

3. Mass of H₂SO₄ = 571.4g

Molecular mass of H₂SO₄ = 98 g/mol

Moles = 571.4g / 98g/mol =5.83 mol

Density = 1.329 g/mL

Weight of solvent = 1.329 g/mL x 1000 mL = 1329 g

Weight of solute = 1329 - 571.4 = 757.6g = 0.757 kg

molality = 5.83 mol / 0.757 kg = 7.70 m

molality = 7.70 m