5) For Trials 2 & 3, the molar mass of the solute was 151 g/mol & 143 g/mol resp

Question

5) For Trials 2 & 3, the molar mass of the solute was 151 g/mol & 143 g/mol respectivctively. A) What is the average molar mass of the solute? B) What are the standard deviation and the relative standard deviation (%RSD) for the molar mass of the solute? B. Freezing Point of Cyclohexane plus Calculation Zone Unknown Solute 2. Mass of cyclohexane (g) 3. Mass of added solute (g) 1014 Part C.4 0.255 C. Calculations I. ky for cyclohexane (C.kg/mol) 2. Freezing point change, 70) 3. Mass of cyclohexane in solution (kg) 4. Moles of solute, total (mol) 20.0 3.04Part C.6 Show calculation. 5, Mass of solute in solution, total og) 6. Molar mass of solute (g/mol) Show calculation. 5. b. For Trials 2 and 3, the molar mass of the solute was 151 g/mol and 143 g/mol respectively a. What is the average molar mass of the solute? b. What are the standard deviation and the relative standard deviation (%RSD) for the molar mass of the solute ?

Explanation / Answer

Trial 1
delta Tf = kf * m => m = (delta Tf)/kf = 3.04/20 = 0.152
molality = (number of moles)/(Mass of solvent) ==> 0.152 = n/0.010.14 ==> n = 0.00154
n = mass/M ==> 0.00154 = 0.255/M ==> M = 165.5844 g

M1 = 165 g
M2 = 151 g
M3 = 143 g

Average molar mass of solute = 153 g

b) SD = (((M-M1)2 + (M-M2)2 + (M-M3)2)/3)1/2  = ((144+4+100)/3)1/2 = 9 g

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