Question #6b36c

So, you need to determine what impact the erroneous measurement of the freezing point of the pure cyclohexane will have on the molar mass of the solute.

The equation for freezing-point depression looks like this

#DeltaT_f = i * K_f * b " "#, where

#DeltaT_f# - the freezing-point depression;
#i# - the van't Hoff factor, equal to #1# for non-electrolytes;
#K_f# - the cryoscopic constant of the solvent;
#b# - the molality of the solution.

Now, the freezing point depression is defined as

#DeltaT_f = T_"f"^0 - T_"f sol"#

Let's assume that you use the correct freezing point of the pure cyclohexane and calculate the molality of the solution to be

#b = (T_"f"^0 - T_"f sol")/(i * K_f)#

The molality of the solution is defined as the moles of solute divided by the mass of the solvent - expressed in kilograms. This means that you have

#n_"solute"/m_"cyclohexane" = (T_"f"^0 - T_"f sol")/(i * K_f)#

#n_"solute 1" = (T_"f"^0 - T_"f sol")/(i * K_f) * m_"cyclohexane"#

Now look what happens when you measure the freezing point of the pure cyclohexane to be #2^@"C"# too high.

#T_"f meas"^0 = T_f^0 + 2^@"C"#

The number of moles of solute will now be

#n_"solute 2" = (T_"f"^0 + 2 - T_"f sol")/(i * K_f) * m_"cyclohexane"#

Since the nominator of the fraction is now bigger than it was when you calculated using the correct value of #T_"f"^0#, it follows that

#(T_"f"^0 + 2 - T_"f sol")/(i * K_f) > (T_"f"^0 - T_"f sol")/(i * K_f)#

#n_"solute 2" > n_"solute 1"#

It now appears that you have more moles of solute in the solution. Since molar mass is defined as mass per moles, you will get

#M_(M 1) = m_"solute"/n_"solute 1" " "# and #" "M_(M 2) = m_"solute"/n_"solute 2"#

Since >#n_"solute 2" > n_"solute 1"#, it follows that

#M_(M 2) < M_(M 1)#

The solute will come out as having a smaller molar mass.