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During a freezing point depression study, two experiments were made with 10 grams of lauric acid....

During a freezing point depression study, two experiments were made with 10 grams of lauric acid. The freezing point of the lauric acid averaged 43.4 deg C. EXP 3: 1.0 grams of an unknown substance are added to the solution and a freezing point for the new solution is 41.0 deg C. EXP 4: 0.5 grams of the same unknown are added to the solution, the new freezing point is 41.6 deg C. Calculate molality of EXP 3 and EXP 4. What are the molar masses of EXP 3 and 4? What is the average molar mass of the unknown substance?

Expert Solution

Freezing point of lauric acid = 43.4 ⁰C

K_f (⁰C Kg/mol) of lauric acid = 3.9 [Source: wikipedia]

For experiment 3,

Freezing point of solution = 41 ⁰C

T_f = T_f (solvent) - T_f (solution) .....(1)

T_f₃ = 43.4 - 41= 2.4 ⁰C

T_f = K_f m ..... (2)

Molality (m₃) = = 0.615 m

Molality =

Molar mass of solute = = 162.60 g/mol

For experiment 4,

Freezing point of solution = 41.6 ⁰C

T_f4 = 43.4 - 41.6 = 1.8 ⁰C [From (1)]

Molality (m₄) = = 0.462 m [From (2)]

Molar mass of solute = = 216.45 g/mol [From (3)]

Average molar mass = = 189.52 g/mol

queen_honey_blossom answered 2 years ago

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