Question

Question #6b36c

Answer 1

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.