# Question 6b36c

##### 1 Answer
Sep 25, 2015

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

$\Delta {T}_{f} = i \cdot {K}_{f} \cdot b \text{ }$, where

$\Delta {T}_{f}$ - 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

$\Delta {T}_{f} = {T}_{\text{f"^0 - T_"f sol}}$

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

$b = \frac{{T}_{\text{f"^0 - T_"f sol}}}{i \cdot {K}_{f}}$

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)#

${n}_{\text{solute 1" = (T_"f"^0 - T_"f sol")/(i * K_f) * m_"cyclohexane}}$

Now look what happens when you measure the freezing point of the pure cyclohexane to be ${2}^{\circ} \text{C}$ too high.

${T}_{\text{f meas"^0 = T_f^0 + 2^@"C}}$

The number of moles of solute will now be

${n}_{\text{solute 2" = (T_"f"^0 + 2 - T_"f sol")/(i * K_f) * m_"cyclohexane}}$

Since the nominator of the fraction is now bigger than it was when you calculated using the correct value of ${T}_{\text{f}}^{0}$, it follows that

$\frac{{T}_{\text{f"^0 + 2 - T_"f sol")/(i * K_f) > (T_"f"^0 - T_"f sol}}}{i \cdot {K}_{f}}$

${n}_{\text{solute 2" > n_"solute 1}}$

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

${M}_{M 1} = {m}_{\text{solute"/n_"solute 1" " }}$ and $\text{ "M_(M 2) = m_"solute"/n_"solute 2}$

Since >${n}_{\text{solute 2" > n_"solute 1}}$, it follows that

${M}_{M 2} < {M}_{M 1}$

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