Question

Benzoic Acid and TBA

Finish the given table:

Trial 1 Trial 2

1. Kf for TBA 9.1° C/m 9.1° C/m

2. Mass of Benzoic Acid (g) 1.01 g 1.01 g

3. Freezing point of benzoic Acid   

and TBA solution from graph 20.4 C 17.6 C

4. Freezing point change 3.8°C 2.6° C

5. Molality of solution ? ?

6. Calculated molar mass of Benzoic Acid (g/mol) ? ?

7. Average Molar Mass (g/mol) ? ?

8. Standard Deviation ?

9. Relative Standard Deviation ?

Answer 1

Trial 1:

molality = mol of solute / kg of solvent

or, form colligative properties

dTf = -Kf*m

where dTf = depression point of mixture

Kf = constant of solvent

m = molality of solvent

so

m = dTf/Kf = 3.8°C/9.1°C/m= 0.41758 molal

Q6.

calculated MM of B.A = mass/mol

we have mass = 1.01g but not moles

mass of solvent --> 19.805 g of solvent

claculate mass

mol of S = molality * kg solvent= (0.41758 )(19.805 /1000) = 0.0082701 mol of S

so

MM = mass / mol = 1.01/0.0082701 = 122.1266 g/mol

Trial 2:

molality = mol of solute / kg of solvent

or, form colligative properties

dTf = -Kf*m

where dTf = depression point of mixture

Kf = constant of solvent

m = molality of solvent

so

m = dTf/Kf = 2.6°C/9.1°C/m= 0.28571 molal

Q6.

calculated MM of B.A = mass/mol

we have mass = 1.01g but not moles

mass of solvent --> 19.805 g of solvent

claculate mass

mol of S = molality * kg solvent= (0.28571)(19.805 /1000) = 0.0056584mol of S

so

MM = mass / mol = 1.01/0.0056584= 178.4 g/mol

Q7.

average...

MW = (178.4+122.1266 )/2 = 150.2633 g/mol

Q8.

STDdev = sqrt(178.4-150.2633)^2 (122.1266 -150.2633g)^2) = 39.79130 g/moll

Std. dev ---> 39.79130/150.2633 = 0.2648 (unitless)