In: Chemistry
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.
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