unknown Substance Benzoic Aud 122.123 g/mol REPORT SHEET Freezing Point Depressi
ID: 530612 • Letter: U
Question
unknown Substance Benzoic Aud 122.123 g/mol REPORT SHEET Freezing Point Depression A. Preparation of experimental a and freezing point determination of laurie acid 1. Mass of freezing point depression apparatus 2. Mass of freezing point depression apparatus with laurie acid 3. Mass of lauric acid used B. Freezing point determination of a laurie acid solution 4. Designation of unknown 5. Mass of added unknown, Exp. 3 6. Mass of added unknown, Exp. 4 Freezing point of lauric acid from Part A, Exp. 1 Freezing point of lauric acid from Part A, Exp.2 Average freezing point of lauric acid Freezing point of lauric acid and unknown from Part B. Exp.3 (Plate) 45. DO Freezing point of lauric acid and unknown from Part B, Exp 4 (tlalo) AT Exp. solution molality (show calculations) Exp.3 0.170 M Exp 4 Moles of unknown (show calculation) Exp 3 Exp 4 mol Molar mass of unknown (show calculation) Exp.3 Exp. 4 M m 1.009 1.30 x 10 mol SSS. 5 g/mol Average molar mass mass of unknown. Show your calculations for the solution molality, mole of unknown, and molar Hand in your cooling curves with your report sheet. sss.S 122.12 Actual Theorencal x 100Explanation / Answer
1) MAJOR SOURCE OF ERRORS ARE: 1)Personal errors: a) error in reading thermometer temperature.
b) using impure reagent
c) Mechanical loss of material
d) error in calculations
2) Operational error: when sound technique is not followed
3) Reagents errors : it may contain impurities
4) Errors in measuring solution.
5) Use of numerically incorrect factors
6) Indeterminant errors : these sre accidental errors in which operator have no control.
3) Colligative properties : those properties which depend upon the number of particle of solute irrespective of their nature relative to the no. of partical present in the solution.So colligative properties of the electrolytic solution will be more then the non electrolytic solution of the same concentration or molality.
Equimolal concentration of any solute will have same freezing point.
Change in temperature = i Kfm (non electrolytes) i =1
Change in temperature = i Kfm (electrolytes)
i = no of dissolved particles.
4) Freezing point = liquid to solid transition.
Increasing order of freezing points ; pure water< sucrose< glucose<NaF < MgI2
freezing point = (no. of ions) (1.86)(molality)
=iKfm (Kf=1.86)
Substract your result from 0 to get freezing point, adding something to water always decreases freezing point.
e.g. change in temperature = 1.25 *1.86*1 =2.33*C
Freezing point = 0 - 2.33 = -2.33*C
6) Strength of the given solution = molality * molar mass
=0.1 *58.5 =5.85 g/L
mass of NaCl in 200g of water = 5.85/ 1000
= 0.00585 *200 = 1.17g
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