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Lab report template for chemistry labs
Typology: Lab Reports
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Experiment # 02 Experiment Title: Colligative Properties Lab report done by: Aaida Hossain - 2132500 Partner: Mia Alyssa Nardelli Presented to: Stephanie Harrison 202-NYA-05, Laboratory Section 15 Date of the Experiment: 09-02- Date of the Lab Report Submission: 23-02-
Results:
Graph 3: The freezing point cyclohexane Graph 4 :The freezing point of the unknown solute/cyclohexane solution
Table 1 - Part A-B: Data for Pure Cyclohexane Mass of empty test tube, stopper and beaker, g 116. Freezing temperature of cyclohexane, ˚C 6. Table 2 - Part A-B: Data for the Benzophenone/Cyclohexane Solution Mass of tube, stopper, beaker and cyclohexane, g 135. Mass of cyclohexane, g 19. Mass of tube, stopper, beaker, benzophenone and cyclohexane, g
Mass od benzophenone, g 0. Freezing Temperature of benzophenone/cyclohexane solution, ˚C
Molality of benzophenone/cyclohexane solution, mol ∙ kg − 1
Kf for cyclohexane, (^) ° C ∙ kg ∙mol−^1 22. Percentage of error, % 13.61% Table 3 - Part A-C: Data for the Pure Cyclohexane Mass of empty test tube, stopper and beaker, g 109. Freezing temperature of cyclohexane, ˚C 6. Kf for cyclohexane from part A-B, (^) ° C ∙ kg ∙mol−^1 22. Table 4 – Part A-C: Data for the Unknown Solute/ Cyclohexane Solution Unknown Number 2 Mass of tube, stopper, beaker and cyclohexane, g 128. Mass of tube, stopper, beaker and unknown solute/cyclohexane solution, g 128. Mass of cyclohexane, g 18. Mass of the unknown solute, g 0. Freezing temperature of unknown solute/ cyclohexane solution, ˚C 4. Molar mass of unknown solute, (^) g ∙ mol−^1 252.
m
∗kg of solvent M (^) unknown solution= 0.3682 g
22.95° C ∙ kg ∙ mol
=252.4 g ∙mol − 1 Pre-laboratory questions:
K (^) f
5.12 kg ∙ °C ∙mol − 1 =1.24^ mol^ ∙^ kg − 1 n=m∗kilogram of solvent =1.24∗0.0100=0.0124 mol
This gives us two equations and two unknowns, allowing us to solve for n and m. This gives n = 3.95 mol and m = 44.2 mol. Molality: molality ( m)= ( moles of solute ) kilogram of solvent
3.95 mol 44.2 mol∗18. g mol ∗ 1 kg 1000 g =4.96 mol / kg Molarity: Molarity= moles of solute liters of solution
3.95 mol 1 L =3.95 mol / L Percent Mass: percent mass= mass of solute mass of solution
3.95 mol∗46. g mol 978 g
References: