Question

Experiment 14 Report Sheet Molar Mass of a Soud 93.4818 TH Part TH Part 3) L. Maaf baket hele 2. Mass of bene Tad man of ded ) Fring point from cooling curve C tractor's approval of graph 97.5819 4 .3378 .444 3 °C 2°C 0°C 1. for cyclohexane C o ) 2. Frais-point charge. ATO 2. Mes of cyclobecane in sosion kg) 4. Males of solte, con mal 5. Mass of solusein solution, anal) 6. Molar mass of solute mal) 7. Average molar mass of solute amo!) Date Analysis, & Standard deviation of molar mass Data Analysis, 9. Relative standard deviation of molar mass (RSD) Dete Analysis, *Show calculation(s) for Trial 2 on the next page. Experiment 14 197

I just need the calculation.

Answer 1

Que to addition of solute in pure sovent, the foreezing point of the solution decreare in freezing point i seint & depression (17) ocur. we know, 47 ak cm. () where, & Cryscopic contant torsolvent Con- Mola concentration of solute here, cyclohexane is solvent. cm. Drumber of mobs of selute man of solvent in kg of man of solute is äg which is dissolved in bg of solvent & Endlas wars of solute à di , then a moles of solute is present & solvent & solvent contain & moles solote in be solven My MX4 MX & bone cyclohexane . Ik gil. 1010 o axlos. a X1000 ot pure cyclohexane Freezing point from cooling curve given = 45°C here, man of solvent - azung ATY = Tf - T' ( = krezept. of pure solvent Freezing son kgr 20 °c tam -

Now, plugging all the values into canation a x1000 ... (i) I T ' = x 2 X Mxte. 45-7 = 20 x a x1000 . (ii) x [: freezing pt. of pane solvent = 45.0 (cfrom cooling norve) k= 20 c kasno be mans of solvent n for trial 1 The gol , au 0.337 g using canation (iii) (45-3) = 20x 0.337 X1000 20 X 8°327x1000 1096 g/mol . Ma 106 g/mol for trial T=2°C, a=0.4449 MX41 Mo 21715 V i vaing (in) 0164 77000 4.5 - 2) = on MX GI .: m= 20x0.444 X1000 2.5x4.1 . M = 866.34 g/mol for frials = 0'c ar oorsag . Erung (ii) as-o = 20 x 8 1659 81000 do 714.36 g/nws Average molar man 1 = [1096 + 866.3477/4.36] glow = 892 23 g i

nul we know standar deviation SD - $(2;-)" where, se number of fince of data i z. saurple size à = Average of the data. xia data point here, n=3, n = 892 123 mil i so 71096 892 257 +(866.34 -892:23) +(71436-892:23) 3-) # - 192.133 Relative standard deviation (RSA) - 4100%. 1921133 X100% 892:23 - 21.534%