Adding solute to water will result in boiling point elevation due to the presence of more molecules. Change in temperature is given by the relation $\displaystyle \bigtriangleup T = k*m*i$, where $\displaystyle k$ is a constant for the solvent, $\displaystyle m$ is the solution molality, and $\displaystyle i$ is the van't Hoff factor. In this example, the molalities are equal.

Since $\displaystyle NaCl$ dissociates into $\displaystyle Na^+$ and $\displaystyle Cl^-$, $\displaystyle i = 2$, representing the two ions derived from each molecule. For glucose $\displaystyle i = 1$, as the molecule does not dissociate. Solution 1 will have a higher elevation in temperature due to the greater number of ions in solution.

Colligative properties are dependent only on the number of particles in a solution, and not their identity. Some examples of colligative properties are vapor pressure, boiling point, freezing point, and osmotic pressure.

There is a direct relationship between the boiling point elevation and the number of particles present in a solution. The more particles that are present in solution, the higher the boiling point elevation.

We are looking for the compound that will create the greatest number of ions when dissolved in solution.

Sodium chloride and magnesium sulfate will produce two ions per mole. $\displaystyle 2mol*\frac{2ions}{1mol}=4\ \text{moles of ions}$

Magnesium chloride and barium chloride will produce three ions per mole. $\displaystyle 1mol*\frac{3ions}{1mol}=3\ \text{moles of ions}$

Calcium hydroxide will also produce three ions per mole, but we are given two moles instead of one. $\displaystyle 2mol*\frac{3ions}{1mol}=6\ \text{moles of ions}$

Calcium hydroxide will produce the greatest number of ions, thus creating the greatest increase in boiling point elevation.

Boiling point elevation is a colligative property, meaning that it depends on the relative number of solute particles in solution. The answer choice with the largest number of moles of particles will show the greatest boiling point elevation. The equation for boiling point elevation is:

$\displaystyle \Delta T_b=k_bmi$

Molality is equal to moles of solute per kilogram of solvent, meaning that it will be proportional to the moles of solute added. Each solute is added to equal amounts of water, allowing us to keep this value constant. Similarly, $\displaystyle \small k_b$ will be constant for all of the solutions. Overall, boiling point elevation will be proportional to the moles of solute multiplied by the van't Hoff factor.

$\displaystyle \Detla T_b\propto ni$

Using this proportion, we can find the solute that will most impact the boiling point of water.

$\displaystyle MgCl_2:\ ni=(1.5)(3)=4.5$

$\displaystyle K_3PO_4:\ ni=(1.25)(4)=5$

$\displaystyle CaCl_2;\ ni=(1.5)(3)=4.5$

$\displaystyle HClO_4:\ ni=(2.5)(2)=5$

$\displaystyle NaCl:\ ni=(3.0)(2)=6$

Since sodium chloride results in the greatest moles of ions in solution, it will yield the greatest boiling point elevation.

In order to solve this problem, we can use the boiling point elevation equation: $\displaystyle \Delta T = k_{b}mi$.

We know the temperature change, we can compute molality from the given information, and we know the van't Hoff factor (expected to be 2 in this scenario due to NaCl becoming 2 ions in solution). We can calculate the boiling point constant for the solvent.

$\displaystyle k_{b}= \frac{\Delta T}{mi}$

$\displaystyle m=\frac{mol}{kg}$

 $\displaystyle k_{b} = \frac{6K}{(\frac{2moles}{1kg})(2)}$

$\displaystyle k_{b} = 1.5\frac{K}{m}$

Recall that the boiling point of a solution depends on the number of ions in the solution and the concentration of the solution. Increase in both of these factors will elevate the boiling point to higher temperatures. The equation describing this is given below.

$\displaystyle \Delta T = k_bim$

where $\displaystyle \Delta T$ is change in boiling point, $\displaystyle k_b$ is the boiling point elevation constant, $\displaystyle i$ is the number of ions, and $\displaystyle m$ is the molality. Sodium chloride, or $\displaystyle NaCl$, will produce two ions in solution whereas calcium chloride, or $\displaystyle CaCl_2$, will produce three ions. If the concentrations were the same, the calcium chloride solution would have a higher boiling point; however, since we are not given the concentration we cannot determine the relative boiling points of the solution.

Recall that adding solutes to a solution increases the boiling point. This phenomenon is called the boiling point elevation. Knowing this information, we can eliminate two choices immediately. The equation to calculate the boiling point elevation is as follows.

$\displaystyle \Delta T = k_bim$

where $\displaystyle \Delta T$ is change in boiling point, $\displaystyle k_b$ is the boiling point elevation constant, $\displaystyle i$ is the number of ions, and $\displaystyle m$ is the molality. The question gives us the concentration in molarity; therefore, we need to convert the concentration to molality.

Molarity = $\displaystyle \frac{moles\: of\: solute}{liters\: of\: solution}$

Molality = $\displaystyle \frac{moles\: of\: solution}{kg\: of\: solvent}$

We need to find the mass of solvent in kilograms. The molarity is $\displaystyle 1M$; therefore, let’s assume we have $\displaystyle 1\:mol$ of solute and $\displaystyle 1L$ of solution. The density of the solution is $\displaystyle 1.5\frac{g}{mL}$ or $\displaystyle 1.5\frac{kg}{L}$. We can calculate the mass of the total solution using density.

mass of total solution = $\displaystyle 1L(\frac{1.5kg}{1L}) = 1.5kg$

We need the mass of the solvent only. To find this, we need to first calculate the mass of solute. The MW of potassium bromide is $\displaystyle 119\frac{g}{mol}$ (this can be calculated by obtaining values from the periodic table). We have $\displaystyle 1\:mole$ of solute; therefore, the mass of solute is

$\displaystyle 1mol\left(\frac{119g}{mol}\right) = 119g = 0.119kg$

This means that the mass of solvent is equal to:

mass of solvent = total mass of solution - mass of solute = $\displaystyle 1.5kg - 0.119kg$ = $\displaystyle 1.381kg$

Molality of solution is

$\displaystyle m = \frac{1\: mol\: solute}{1.382\: kg\: solvent} = 0.724m$

$\displaystyle i$, or the number of ions, for $\displaystyle KBr$ is 2 (because $\displaystyle KBr$ will produce two ions in solution). Now we have all the information to calculate the boiling point elevation.

$\displaystyle \Delta T = \frac{0.512^oC}{m}(2)(0.724m) = 0.741^oC$

Therefore, boiling point increases by $\displaystyle 0.741^oC$.

We are looking for the least amount of freezing point depression. Freezing point depression is calculated using the equation:

$\displaystyle \Delta T_{f} = k_{f}im$

Each of these solutions has a different molality, which needs to be calculated. Molality is equal to moles of solute per kilogram solvent. To find this value, convert grams to moles (using molar mass) and divide by the mass of the solvent.

$\displaystyle m=\frac{g*\frac{mol}{g}} {0.5kg\ H_2O}$

$\displaystyle Ba(OH)_2:\ \frac{(10g)(\frac{1mol}{171.3g})}{0.5kg}=0.117m$

$\displaystyle C_6H_{12}O_6:\ \frac{(40g)(\frac{1mol}{180g})}{0.5kg}=0.444m$

$\displaystyle K_2O:\ \frac{(50g)(\frac{1mol}{94g})}{0.5kg}=1.06m$

$\displaystyle NaCl:\ \frac{(25g)(\frac{1mol}{58.5g})}{0.5kg}=0.855m$

Next, find the van't Hoff factor for each compound.

$\displaystyle Ba(OH)_2:\ i=3$

$\displaystyle C_6H_{12}O_6:\ i=1$

$\displaystyle K_2O:\ i=3$

$\displaystyle NaCl:\ i=2$

Finally, use the initial equation to find the smallest freezing point depression (we are looking for the solution with the highest freezing point).

$\displaystyle \Delta T_{f} = k_{f}im$

$\displaystyle Ba(OH)_2:\ (1.86\frac{^oC}{m})(3)(0.117m)=0.65^oC$

$\displaystyle C_6H_{12}O_6:\ (1.86\frac{^oC}{m})(1)(0.444m)=0.83^oC$

$\displaystyle K_2O:\ (1.86\frac{^oC}{m})(3)(1.06m)=5.93^oC$

$\displaystyle NaCl:\ (1.86\frac{^oC}{m})(2)(0.855m)=3.18^oC$

After calculating the change in freezing point for each solution, we find that the barium hydroxide solution has the smallest depression.  

First, calculate the freezing point depression with the equation:

$\displaystyle \Delta T_{f} = k_{f}im$

The van't Hoff factor for glucose is 1, since it does not dissociate in solution. The freezing point depression constant is given in the table.

$\displaystyle \Delta T_f=(4.90\frac{^oC}{m})(1)(0.2m)$

$\displaystyle \Delta T_f=0.98^oC$

Next, subtract this value from the freezing point of pure benzene to find the freezing point of the final solution.

$\displaystyle 5.5^oC-0.98^oC=4.52^oC$

Freezing point depression is given by the equation $\displaystyle \Delta T_{f} = -k_{f}im$.

i is the Vant Hoff Factor that tells us how many particles (often ions) a solid produces when dissolved in solution. m is the molality of the solution. k_f is the freezing point constant.

The question gives us m as 3mol, k_f as 1.9, and i is found to be 3 (2K⁺ and SO₄^2- ). Whe then plug the values into the equation.

$\displaystyle \Delta T_{f} = -(1.9\frac{^oC}{mol})(3)(3mol) = -17.1^oC$

Freezing point depression is a colligative property, meaning that it depends on the amount of particles present in solution. The more ions that a solute dissociates into, the larger the magnitude of freezing point depression. Here, we're looking for the smallest amount of depression, so we want to find the solute that dissociates into the fewest particles. The answer is glucose, which is not ionic and does not dissociate at all in solution.