Question

1.0 gram samples of solutes A,B, and C are dissolved in 10.00 g of water to give three separate solutions. Use the information below to calculate the molecular mass of A, B, and C. Determine the molecular formula of solution B. Assume all solutes are nonvolatile and nonelectrolytes. 1. Solute A has a vapor pressure of 16.9 torr at 20 degrees C. The Vapor pressure of water at 20 degrees C si 17.5 torr. Calculate Molecular mass. 2. Solute B freezes at -1.03 degrees C and has an analysis of 40% Carbon, 53.6% Oxygen, and 6.7% Hydrogen. The Kf (constant of Freezingpoint) for water is 1.86. What is the moleuclar mass and molecular formula? 3. Solute C has an osmotic pressure of 1.6 atm at 25 degrees C. What is the molecular mass of this substance?

Answer 1

1. Sample A : 1 g in 10 g water

mole fraction of water = 16.9 torr/17.5 torr = 0.966

mole fraction of A = 1-0.966 = 0.034

moles fraction of water = moles of water/total moles

moles of water = 10/18 = 0.55 mols

total moles = 0.55/0.966 = 0.57 mols

moles of A = 0.57 - 0.55 = 0.02 mols

molecular mass of A = 1/0.02 = 50 g/mol

2. Sample B 1 g in 10 g water

dTf = -1.03 = -1.86 x m

molality (m) = 0.55 m

molecular mass of B = 1/0.55 x (10/1000) = 181.82 g/mol

C = 40% ; moles of C = 40/12 = 3.33 mols

H = 6.7% ; moles of H = 6.7/1 = 6.7 mols

O = 53.6% ; moles of O = 53.6/16 = 3.35 mols

divide by smallest fraction,

C = 3.33/3.33 = 1

H = 6.7/3.33 = 2

O = 3.35/3.33 = 1

empirical formula = CH2O

empirical formula mass = 12 + 2 + 16 = 30 g

ratio = molecular mass/empirical formula mass = 181.82/30 = 6

molecular formula of B = C6H12O6

3. Solute C 1 g in 10 g water

osmotic pressure = 1.6 atm = MRT

1.6 atm = M x 0.08206 x (273 + 25)

molarity = 0.065 M

molecular mass of C = 1/0.065 x 0.01 = 1538.46 g/mol