The student added 1.85 g of an unknown molecule to the same sample of benzopheno

Question

The student added 1.85 g of an unknown molecule to the same sample of benzophenone used in figure 2. The sample is heated above the freezing point then cooled and the temperature of the sample was plotted as a function of time (Figure 3). A. Show your extrapolation work on Figure 3 and record the freezing point below. B. Determine the freezing point depression of the solution (AT). Show your work below. Include units. C. Determine the molality (m) of the solution used in Figure 3. Show detailed work below. Include units. D. Determine the # mol of the unknown solute used in problem 2. Show your work below. Include units. E. Determine the molar mass of the unknown solute used in problem 2. Show your work below. Include units.

Explanation / Answer

(A)

From the plot given, the freezing point of the solution can be approximated as 390C.

(B)

Freezing point of pure benzophenone = 48.50C

So, depression in freezing point, dT = 39-48.5 = -9.50C

(C)

For benzophenone, Kf = 9.8 C/m

Using equation:

dT = -Kf * m

So, m = -(-9.5)/9.8 = 0.97 molal

(D)

Since you have not mentioned the mass of benzophenone solvent taken, I assume it to be 1 kg

Now,

m = moles of solute / mass of solvent in gs

Thus,

#moles of solute = 0.97*1 = 0.97 moles

(E)

MW of unknown = mass taken / moles = 1.85/0.97 = 1.9 g/mol

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