A student determines the freezing point of a solution of 0.63 g of mandelic acid

Question

A student determines the freezing point of a solution of 0.63 g of mandelic acid in 20.78 g of tertiary butyl alcohol. He obtains the following temperature-time readings: Plot these data on the graph paper provided. Note that the first several points fall roughly on a straight line. The last several points fall on a better straight line. Draw in those two straight lines as best you can. The point at which the lines intersect is the freezing point of the solution. Ignore the rather substantial supercooling that occurred What is the freezing point of the solution? What is the freezing point depression, T_t^0 - T_f, or Delta T_t? Take _t^0T to be 24.5 degree C for TBA. What is the molality m of the mandelic acid in the solution? (Use Eq. 1: k_f = 809.) What is the molar mass of mandelic acid? Use Equation 2 as modified below: MM = no. g solute/no. kg solvent times 1/m The molecular formula of mandelic acid is C_6H_5CH (OH)-COOH. Is the result you obtained in Part e consistent with this formula? Any comments?

Explanation / Answer

1 a. Plot the graph first.

b. The freezing point of the solution is obtained from the intersection of the two straight lines. Let at time t, the freezing point be m.

Plug these two unknowns in the two equations above:

m = -4.0405t + 34.258

m = -0.2467t + 23.313

Since these two values are equal,

-4.0405t + 34.258 = -0.2467t + 23.313

===> 34.258 – 23.313 = -0.2467t + 4.0405t

===> 10.945 = 3.7938t

===> t = 10.945/3.7938 = 2.88497 2.8850

At 2.8850 minutes, the two straight lines will meet. Put the value of t in either of the two equations to obtain m.

m = -0.2647*2.8850 + 23.313 = 22.549 22.55

The freezing point of the solution is 22.55C (ans).

c. The freezing point depression is given as Tt = Tf - Tt

Given Tf = 24.5C and Tt = 22.55C, we have the freezing point depression, Tf = (24.5 – 22.55)C = 1.95C (ans).

d. As per Raoult’s law, we must have,

Tf = kf. m where m = molality of the solution; kf = 8.09 C/molal

Plug in values to obtain

1.95C = (8.09C/molal).m

===> m = 1.95/8.09 molal = 0.241 molal (ans).

e. Use the given formula to obtain the molecular mass of madelic acid as

MM = (no. g solute/no. kg solvent)*1/m

===> MM = (0.63 g)/[(20.78 g)*(1 kg/1000 g)]*1/0.241

===> MM = (0.63*1000)/(20.78)*1/0.241 = 125.799 125.8

The MM of mandelic acid is 125.8 (ans).

f. The molecular formula for madelic acid is C6H5CH(OH)COOH. Noting the atomic masses of the elements as

C = 12

H = 1

O = 16

calculate the formula MM as (8*12 + 8*1 + 3*16) = 152 (ans).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.