A nonvolatile solute will not contribute to the vapor pressure of the solution, but will decrease the vapor pressure of the pure solvent. We can solve for the vapor pressure of the solution by using Raoult's law:

 $\displaystyle P_v=X_aP$

In other words, the new vapor pressure is equal to the molar fraction of the solvent multiplied by the vapor pressure of the pure solvent.

$\displaystyle P_{v} = (\frac{8moles}{10 moles})(760mmHg) = 608mmHg$

Since both of the solvents have a vapor pressure, we can find the percentage of ethanol in the solution by using Raoult's law, including both solvents in the equation:

$\displaystyle P_v=X_aP_a+X_bP_b$

Since we want to find the percentage of ethanol in the solution, we will designate the molar fraction of ethanol as $\displaystyle X$. Since the sum of the molar fractions equals one, the molar fraction of hexane will be designated as $\displaystyle X-1$.

$\displaystyle 80mmHg = (X)(50mmHg) + (1-X)(132mmHg)$

$\displaystyle 80mmHg=X(50mmHg)+132mmHg-X(132mmHg)$

$\displaystyle -52mmHg=X(-82mmHg)$

$\displaystyle X=0.634=63.4\%$

So, 63.4% of the solution is ethanol.

Since the solute is volatile, it will contribute to the total vapor pressure of the solution. As a result, we must incorporate it when solving for the total vapor pressure.

$\displaystyle P_v=P_{va}+P_{vb}$

We can find the vapor pressures using Raoult's law.

$\displaystyle P_{va}=X_aP_a$

$\displaystyle P_{vb}=X_bP_b$

$\displaystyle \small P_{v} = X_{a}P_{a} + X_{b}P_{b}$

Since we are solving for the molar fraction of the solvent, we will designate its molar fraction as $\displaystyle \small X$. The sum of the molar fractions of each component must be equal to 1; thus, the molar fraction of the solute must be $\displaystyle \small 1-X$. Using these variables and the information given, we can solve for the molar fraction of the solvent.

$\displaystyle \small 68mmHg = (X)(85mmHg) + (1-X)(23mmHg)$

$\displaystyle 68mmHg=X(85mmHg)+23mmHg-X(23mmHg)$

$\displaystyle 45mmHg=X(62mmHg)$

$\displaystyle \small X = 0.726$

The equation for freezing point depression is $\displaystyle \Delta T_F = k_f m i$, where $\displaystyle \Delta T_F$ is the change in temperature, $\displaystyle k_f$ is a constant related to the solvent, $\displaystyle m$ is molality, and $\displaystyle i$ is the van't Hoff factor, which is the number of ion particles from each dissolved molecule. We simply plug these numbers into the equation to find the new freezing point.

We know our constant is $\displaystyle \small k_f_{water}=1.853\frac{C^okg}{mol}$. Molality is moles of solute per kilogram of solution, and we know that the density of water is one kilogram per liter and the molecular weight of sodium chloride is $\displaystyle \small 58.5\frac{g}{mol}$.

$\displaystyle 2L*\frac{1kg}{1L}=2kg\ H_2O$

$\displaystyle 174g*\frac{1mol}{58.5g}=2.97mol\ NaCl$

The van't Hoff factor is $\displaystyle \small 2$. Sodium chloride creates only two ions when dissolved, one $\displaystyle Na^+$ and one $\displaystyle Cl^-$.

Using these values, we can solve for the freezing point depression.

$\displaystyle \Delta T_F = (1.853\frac{C^okg}{mol})(\frac{2.97mol}{2kg})(2)$

$\displaystyle \Delta T_F=(1.853\frac{C^okg}{mol})(1.49\frac{kg}{mol})(2)=5.51C^o$

The freezing point will be decreased by $\displaystyle \small 5.51C^o$. The normal freezing point is $\displaystyle \small 0C^o$, making the new freezing point $\displaystyle \small -5.51C^o$.

Since a salt has been added to the pure water, we can find the new melting point of the solution by using the freezing point depression equation:

$\displaystyle \Delta T=k_fmi$

The change in temperature is equal to the freezing point constant for the solvent multiplied by the molality of the solution multiplied by the van't Hoff factor. The van't Hoff factor is the number of ions that a salt will dissociate into when in solution.

$\displaystyle MgCl_2\rightarrow Mg^{2+}+2Cl^-$

For this particular salt, the van't Hoff factor is three. The molality will be equal to the moles of solute over the mass of the solvent.

$\displaystyle 3L\ H_2O*\frac{1000mL}{1L}*\frac{1mg}{1mL}=3000g\ H_2O=3kg\ H_2O$

$\displaystyle 100g\ MgCl_2*\frac{1mol}{95.2g}=1.05mol\ MgCl_2$

Using these terms together in the original equation, we can find the freezing point depression.

$\displaystyle \Delta T=(1.86)(\frac{1.05mol}{3kg})(3)$

$\displaystyle \Delta T=1.95^oC$

This is the change in temperature from the regular freezing point. Since the freezing point for pure water is $\displaystyle 0^oC$, the new melting point is $\displaystyle -1.95^{\circ}C$.