Stearic acid melts at 69.3 °C and has a freezing point depression constant of 4.

Question

Stearic acid melts at 69.3 °C and has a freezing point depression constant of 4.50 °C kg/mol. Imagine that when doing your experiment, you accurately measure the freezing point of stearic acid (69.3 °C) and you obtain a solution freezing point of 66.2 °C for the solution of the stearic acid with your unknown. If you had used 9.0000 g of stearic acid and 0.7500 g of your unknown, what would the molar mass of the solute be? 2. 3. In the event you accidentally add 1.5000 g of unknown solid to the test tube instead of the 0.75 g instructed, how will the freezing point of the solution be affected? Will this change your ability to identify the unknown?

Explanation / Answer

2) dTf = i*Kf*m

DTf = T0-Ts = 69.3-66.2

i= vanthoff factor of solute = 1

Kf of stearic acid = 4.5 c/m

   m = molality = (w/M)*(1000/W in g)

                = (0.75/x)*(1000/9.0)

(69.3-66.2) = 1*4.5*(0.75/x)*(1000/9.0)

x = molarmass of unknown = 121 g/mol

3)

if w = 1.5

(69.3-x) = 1*4.5*(1.5/121)*(1000/9.0)

x = freezing point of solution = 63.1 c

freezing point of solution decreases.

no it wont changes the ability. if freezing point measures correctly.

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