A) If the freezing point depression constant for camphor is 40°C·kg/mol, determi

Question

A) If the freezing point depression constant for camphor is 40°C·kg/mol, determine the molal concentration of isoborneol in the product given that the melting point of your camphor sample is 157°C and the true melting point of camphor is 179°C.

B) Given the molality of isoborneol is 0.340 mol/kg, what is the percent by mass of isoborneol in the product?

C) Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that the melting point of pure camphor is 179°C and its freezing point depression constant is 40°C·kg/mol?

D) Assume you begin with 9.2 mmol of camphor. What is your percent yield if your experimental yield is 0.045 g?

Explanation / Answer

K = 40°m

find m of isoborneaol

T = 157°C

Tf = 179°C

dT = K*m

(179-157) = 40*m

m = (179-157)/40 = 0.55 mol/kg

b)

m = 0.34 mol/kg

find % by mass

MW of isoborneol = 154.25

mass = mol*mw = 0.34*154.25 = 52.445 g

MT = 52.445 + 1kg = 1.052445

%M = 52.445/1052.45 *100 = 4.983%

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