# Question #6b36c

##### 1 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

*van't Hoff factor*, equal to

**cryoscopic constant** of the solvent;

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 *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 + 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 >

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

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