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.

Freezing-point depression is a colligative property and is calculated with the formula:

\(\displaystyle \Delta T_f=-k_fim\)

The van't Hoff factor, \(\displaystyle i\), represents the number of particles a compound will dissociate into when dissolving in a solution. \(\displaystyle k_f\) is the freezing-point depression constant, which is given in the question. \(\displaystyle m\) is molality.

Using the equation, we can calculate the freezing point depression using the values from the question.

\(\displaystyle \Delta T_f=-(1.9\frac{^\circ C*kg}{mol})(2 m) (3)\)

\(\displaystyle \Delta T_f=-11.4^\circ C\)

The standard freezing point of water is \(\displaystyle \small 0^oC\). With the freezing point depression, the solution will freeze at \(\displaystyle 0^oC-11.4^oC=-11.4^oC\).

Freezing-point depression is a colligative property and is calculated with the formula:

\(\displaystyle \Delta T_f=-k_fim\)

The van't Hoff factor, \(\displaystyle i\), represents the number of particles a compound will dissociate into when dissolving in a solution. \(\displaystyle k_f\) is the freezing-point depression constant, which is given in the question. \(\displaystyle m\) is molality.

Using the equation, we can calculate the freezing point depression using the values from the question.

\(\displaystyle \Delta T_f=-(1.9\frac{^\circ C*kg}{mol})(2 m) (2)\)

\(\displaystyle \Delta T_f=-7.6^oC\)

The standard freezing point of water is \(\displaystyle \small 0^oC\). With the freezing point depression, the solution will freeze at \(\displaystyle 0^oC-7.6^oC=-7.6^oC\)

The key to understanding this problem is realizing that it is asking about freezing point depression, or colligative properties. The main factor that concerns freezing point depression is the number of particles in solution, rather than other features, such as size or weight. The more solute dissolved, the greater the freezing point depression will be, meaning that a colder temperature will be needed for a solution to freeze.

In our answer choices we are looking for the solute that is dissociate into the greatest number of ions, or has the highest van't Hoff factor. Magnesium chloride and calcium bromide both generate three moles of ions per mole of solute. The answer choice with magnesium chloride is more concentrated, however, making this the correct option.