Question #6b36c

1 Answer
Sep 25, 2015

Answer:

The molar mass of the solute will come out to be smaller than it actually is.

Explanation:

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.