Irial 2 ol test tube Mass of cyclohexane Freezing point of pure cyclohexane Mass
ID: 695642 • Letter: I
Question
Irial 2 ol test tube Mass of cyclohexane Freezing point of pure cyclohexane Mass of naphthalene and paper Mass of paper Total mass of naphthalene in solution 19 315 Freezing point of solution Freezing-point depression, AT Molality of solution, m Molal freezing-point constant Average k B. Molecular Weight of an Unknown Solute /12 Trial 2 Trial 1 10 7.83) Mass of test tube and cyclohexane Mass of test tube Mass of solvent Mass of sample and paper Mass of paper Total mass of solid sample in solution Freezing point of solvent (from part A) Freezing point of solution Freezing-point depression, ar, Apparent molality of solution Grams of solute per kg of solvent Molecular weight of solute (grams per mole) Average molecular weight of solute 92 611) O , 3 3039- 129Explanation / Answer
Trial 1
Trail 2
Mass of test tube and cyclohexane (g)
107.83
107.83
Mass of test tube (g)
97.694
97.694
Mass of solvent (g)
10.136
10.136
Mass of sample and paper (g)
4.39
4.39
Mass of paper (g)
1.36
1.36
Total mass of solid sample in solution (g)
3.03
3.03
Freezing point of solvent (from part A) (°C)
7 (you have noted the freezing point of pure cyclohexane in part A as 7°C)
7
Freezing point of solution (°C)
3
3
Freezing point depression, Tf = (freezing point of pure solvent) – (freezing point of solution) (°C)
4
4
Apparent molality of solution (m)
Use the relation Tf = kf*m where kf = freezing point depression constant of the pure solvent and m = molality of the solution.
For cyclohexane, we have, kf = 20.0°C.kg/mol; plug in values and obtain
4°C = (20.0°C.kg/mol)*m
====> m = 0.2 mol/kg
Grams of solute per kg of solvent (g/kg)
Mass of solvent taken = 10.136 g = (10.136 g)*(1 kg/1000 g) = 0.010136 kg.
Grams of solute per kg of solvent = (3.03 g)/(0.010136 kg) = 298.934 g/kg
Molecular weight of solute (g/mol)
(Grams of solute per kg of solvent)/(molality of solution) = (298.934 g/kg)/(0.2 mol/kg) = 1494.67 g/mol
Average molecular weight of solute (g/mol)
1494.67 g/mol
Trial 1
Trail 2
Mass of test tube and cyclohexane (g)
107.83
107.83
Mass of test tube (g)
97.694
97.694
Mass of solvent (g)
10.136
10.136
Mass of sample and paper (g)
4.39
4.39
Mass of paper (g)
1.36
1.36
Total mass of solid sample in solution (g)
3.03
3.03
Freezing point of solvent (from part A) (°C)
7 (you have noted the freezing point of pure cyclohexane in part A as 7°C)
7
Freezing point of solution (°C)
3
3
Freezing point depression, Tf = (freezing point of pure solvent) – (freezing point of solution) (°C)
4
4
Apparent molality of solution (m)
Use the relation Tf = kf*m where kf = freezing point depression constant of the pure solvent and m = molality of the solution.
For cyclohexane, we have, kf = 20.0°C.kg/mol; plug in values and obtain
4°C = (20.0°C.kg/mol)*m
====> m = 0.2 mol/kg
Grams of solute per kg of solvent (g/kg)
Mass of solvent taken = 10.136 g = (10.136 g)*(1 kg/1000 g) = 0.010136 kg.
Grams of solute per kg of solvent = (3.03 g)/(0.010136 kg) = 298.934 g/kg
Molecular weight of solute (g/mol)
(Grams of solute per kg of solvent)/(molality of solution) = (298.934 g/kg)/(0.2 mol/kg) = 1494.67 g/mol
Average molecular weight of solute (g/mol)
1494.67 g/mol
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