6. A CHM 126 student followed the procedure t\'s data table below or benzenedepr

ID: 551372 • Letter: 6

Question

6. A CHM 126 student followed the procedure t's data table below or benzenedepression. Fill in the blanks in the studehttical report correct number of figures Molar mass by freezing p and determi outlined for the experiment: Determination of the ental value of molar mass of benzene, CHo the t error in the measurement. Be to ene and percent error in the measurement. re to use proper rounding procedures and sgnificant figures in the values in the table. (10 points) 15.1565 °C/m Mass of Test Tube and Solvent he Freezing Point Depression Constant Ck) for solvent 22.4265 p Mass of Empty Test Tube 6.1254 g Mass of Solvent Freezing pointofPure Solvent in Run #1 Mass of Benzene used inRun #2 Freezing Point of Solution in Run #2 5.128 °C 1.0086 8 -8.168 oC Experimental Molar Mass of Benzene (CH) Theoretical Molar Mass of Benzene (CH) Percent Error in Experimental Molar Mass of Benzene (C.H) ow your work here:

Explanation / Answer

Value of Freezing Point Depression Constant (Kf) for solvent

15.1565 °C/m

Mass of test tube and solvent

22.4265 g

Mass of empty test tube

6.1254 g

Mass of solvent

(22.4265 – 6.1254) g = 16.3011 g

Freezing point of pure solvent in run #1

5.128 °C

Mass of benzene used in run #2

1.0086 g

Freezing point of solution in run #2

-8.168°C

Experimental Molar Mass of Benzene (C6H6)

70.531 (check calculation 1)

Theoretical Molar Mass of Benzene (C6H6)

(6*12.01 + 6&*1.008) = 78.108 au

Percent Error in experimental molar mass of Benzene (C6H6)

(78.108 - 70.531 au)/(78.108 au)*100 = 9.70%

Calculation 1:

Mass of solvent = 16.3011 g = (16.3011 g)*(1 kg/1000 g) = 0.0163011 kg.

Depression in freezing point of the solvent = T = Tpure – Tsolution = (5.128 °C) – (-8.168°C) = 13.296°C.

Let the molecular mass of benzene be M; therefore, mole(s) of benzene taken = (1.0086 g/M).

Molarity of the solution = (moles of Benzene)/(kilograms of solvent) = (1.0086 g/M)/(0.0163011 kg) = (1.0086)/(0.0163011) m

By the problem,

T = Kf*m

====> 13.296°C = (15.1565 °C/m)*(1.0086/M)/(0.0163011) m

====> 13.296°C = (15.1565 °C)*(1.0086/0.0163011.M)

====> M = (15.1565)*(1.0086)/(13.296).(0.163011) = 70.531

The theoretical molar mass of Benzene is 70.531 au.