Boiling Point

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MCAT Physical › Boiling Point

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Colligative properties are properties of compounds that are altered by the amount of substance present. There are four main colligative properties: boiling point, freezing point, vapor pressure, and osmotic pressure. The change in each of these properties can be calculated using the amount of molecules/ions present in solution and the concentration or partial pressure of the compound. The boiling point is defined as the temperature at which the vapor pressure equals the atmospheric pressure. The freezing point is the temperature at which a liquid is converted to a solid. Vapor pressure is the pressure produced by the vapor above a solution. Osmotic pressure is the pressure required to prevent flow of water into a solution (across a membrane).

Which of the following is true regarding the boiling point of a sodium chloride solution and calcium chloride solution?

The relative boiling points cannot be determined because the concentration is not given

The boiling point of calcium chloride solution will be higher

The boiling point of sodium chloride solution will be higher

The boiling point of both solutions will be equal

Explanation

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.

$\Delta T = k_bim$

where $\Delta T$ is change in boiling point, is the boiling point elevation constant, is the number of ions, and is the molality. Sodium chloride, or , will produce two ions in solution whereas calcium chloride, or , 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.

Which of the following compounds will create the greatest increase in boiling point when added to an aqueous solution?

$2 mol\ Ca(OH)_2$

$2mol\ NaCl$

$1mol\ MgCl_2$

$2mol\ MgSO_4$

$1mol\ BaCl_2$

Explanation

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. $2mol*\frac{2ions}{1mol}=4\ \text{moles of ions}$

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

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

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

Boiling point is the temperature a liquid needs to achieve in order to begin its transformation into a gaseous state. Campers and hikers who prepare food during their trips have to account for differences in atmospheric pressure as they ascend in elevation. During the ascent, the decrease in atmospheric pressure changes the temperature at which water boils.

Further complicating the matter is the observation that addition of a solute to a pure liquid also changes the boiling point. Raoult’s Law can be used to understand the changes in boiling point if a non-volatile solute is present, as expressed here.

$\LARGE P = X_{solv} * P_{solv}$

In this law, $\small X_{solv}$ is the mole fraction of the solvent, $\small P_{solv}$ is the vapor pressure of the pure solvent, and is the vapor pressure of the solution. When this vapor pressure is equal to the local atmospheric pressure, the solution boils.

The change in boiling point with addition of a solute is a colligative property of a solution. Which of the following are also examples of colligative properties?

I. Vapor pressure reduction

II. Color emission with dissolution of a solute

III. Osmotic pressure

I and III, only

I, II, and III

III, only

II and III, only

I and II, only

Explanation

Colligative properties are defined as properties that depend entirely upon the ratio of the number of solute particles to the number of solvent particles. Only osmotic pressure and vapor pressure depression are examples of such phenomena. While color emission is a property of a solution, it depends on the chemical species involved, and not the number of particles.

The values for normal boiling and freezing points, along with $\small k_b$ and $\small k_f$ values are given below for select solvents.

$\begin{matrix} \text{Solvent} & \text{Boiling Point}\ (^oC) & k_b\ (\frac{^oC}{m}) & \text{Freezing Point}\ (^oC) & k_f\ \frac{^oC}{m}\ \text{Benzene} & 80.1 & 2.53 &5.5 &4.90 \ \text{Water}& 100.0 & 0.52 & 0.0 & 1.86\ \text{Acetic Acid} & 117.9 & 3.07 & 16.6 & 3.90 \end{matrix}$

$\small 25g$ of ammonium phosphate are dissolved in $\small 500mL$ of water. What is the boiling point of this solution at $\small 1atm$ ?

Explanation

We first need to find the boiling point elevation with the equation:

$\Delta T_{b}=k_{b}im$

Ammonium phosphate has an van't Hoff value of four; each molecule dissociates into four ions in solution. To calculate the molality, we need to find moles of solute per kilogram of solution.
$25 g\ (NH_{4})_{3}PO_{4} * \frac{1 mol}{149g} = 0.17 mol\ (NH_4)_3PO_4$

$500mL\ H_2O*\frac{1g}{1mL}=500g=0.5kg\ H_2O$

$m=\frac{mol\ (NH_4)_3PO_4}{kg\ H_2O}=\frac{0.17mol}{0.5kg}=0.336m$

Next, use the molality, van't Hoff factor, and boiling point elevation constant to solve for the increase in boiling point.

$\Delta T_{b}=k_{b}im$

$\Delta T_b=(0.52\frac{kg^oC}{mol})(4)(0.336m)$

$\Delta T_b=0.70^oC$

Add this increase to the boiling point of pure water to find the boiling point of the solution.

Two moles of sodium chloride (NaCl) are added to 1kg of a mystery solvent. The addition of the NaCl caused an increase of 6K to the solvent's boiling point.

Based on this information, what is the boiling constant for the solvent?

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

$k_{b} = 3.0\frac{K}{m}$

$k_{b}= 6.0\frac{K}{m}$

$k_{b}= 1.0\frac{K}{m}$

Explanation

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

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.

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

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

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

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

Which of the following aqueous solutions will have the highest boiling point?

2m MgCl2

1m MgCl2

2m NaCl

1m NaF

Explanation

In order to answer this problem, consider the equation for boiling point elevation: $\Delta T = k_{b}mi$ .

$k_{b}$ is a specific constant for the boiling substance, so it will not change between the solutions (they are all aqueous). Molality is designated as "m", and a high molality will result in a higher boiling point, however, the value we want to look at for this problem is , which is also known as the van't Hoff factor. The van't Hoff factor is the number of particles that a single solute will dissociate into when added to a solution. MgCl2 will dissociate into three particles: 1 Mg2+ cation and 2 Cl- anions. Since 2m of MgCl2 has the highest molality as well as the largest van't Hoff factor out of the options, it will result in the highest boiling point.

Each of the following solutions is added to equal amounts of water. Which solution will result in the greatest amount of boiling point elevation?

$3.0mol\ \text{of}\ \text{NaCl}$

$1.5mol\ \text{of}\ \text{CaCl}_2$

$1.25mol\ \text{of}\ \text{K}_3\text{PO}_4$

$1.5mol\ \text{of}\ \text{MgCl}_2$

$2.5mol\ \text{of}\ \text{HClO}_4$

Explanation

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:

$\Delta T_b=k_bmi$

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, $\small k_b$ 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.

$\Detla T_b\propto ni$

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

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

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

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

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

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

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

Which solution will have a higher boiling point?

Solution 1: in water

Solution 2: in water

Solution 1

Solution 2

Solution 1 and 2 will have the same boiling point

The answer cannot be determined from the information given

Explanation

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 $\bigtriangleup T = k*m*i$ , where is a constant for the solvent, is the solution molality, and is the van't Hoff factor. In this example, the molalities are equal.

Since dissociates into and , , representing the two ions derived from each molecule. For glucose , as the molecule does not dissociate. Solution 1 will have a higher elevation in temperature due to the greater number of ions in solution.

The values for normal boiling and freezing points, along with $\small k_b$ and $\small k_f$ values are given below for select solvents.

A $\small 0.05m$ solution of which of the following compositions would result in the greatest boiling point elevation?

Magnesium phosphide in acetic acid

Sodium chloride in benzene

Sodium chloride in acetic acid

Magnesium phosphide in benzene

Explanation

Boiling point elevation depends on three variables: the boiling point elevation constant of the solvent, the van't Hoff factor of the solute, and the molality of the solution. In this question, molality is held constant.

$\Delta T_{b} = k_{b}im$

First, calculate the van't Hoff for each compound.

$Mg_3P_2\rightarrow i=5$

$NaCl\rightarrow i=2$

Ultimately, we are looking for the greatest product of the boiling point elevation constant and van't Hoff factor (since molality is constant).

$\text{Benzene}\rightarrow k_b=2.53\frac{^oC}{m}$

$\text{Acetic Acid}\rightarrow k_b=3.07\frac{^oC}{m}$

Magnesium phosphide has the greater van't Hoff factor and acetic acid has the greater boiling point elevation constant. A solution of magnesium phosphide in acetic acid will thus have the greatest boiling point elevation.

$\Delta T_b=(3.07\frac{^oC}{m})(5)(0.05m)=0.77^oC$

Suppose two containers each contain the same amount of solvent. 2m NaCl solution is added to the first container, and a mystery solution is added to the second container. Upon heating the flasks, it is determined that the second container has a higher boiling point than the first container. Assume the solutions are ideal.

Based on the above information, which of the following compounds could have been added to container 2?

2m CaF2

1m MgCl2

3m C6H12O6

1m NaCl

Explanation

Based on the equation $\Delta T_{b} = k_{b}mi$ , we see that there are two factors that differ between the containers and can affect the elevation of the boiling point: molality and the van't Hoff factor ().

The sodium choride added to container 1 has a molality of 2, as well as a van't Hoff factor of 2. As a result, we are looking for a compound that has a larger combination of these two factors, which would cause a higher boiling point. 2m CaF2 has a molality of 2 and a van't Hoff factor of 3. Since this combination of factors in container 2 would be higher than the combination in container 1, we can conclude that this was the mystery compound added to the container with the higher boiling point.

Note that C6H12O6 is the formula for glucose, and will not ionize in solution.

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