Find the Boiling Point of the Following Solutions 2.00 M

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Which solution will have a higher boiling point?

Solution 1: 100mol NaCl in water

Solution 2: 100mol Glucose in water

Possible Answers:

Solution 1

Solution 1 and 2 will have the same boiling point

The answer cannot be determined from the information given

Solution 2

Correct answer:

Solution 1

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

Correct answer:

$2 mol\ Ca(OH)_2$

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.

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

Correct answer:

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

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 ofparticles 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.

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?

Correct answer:

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

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.

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

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

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

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

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?

Possible Answers:

The boiling point of calcium chloride solution will be higher

The boiling point of both solutions will be equal

The boiling point of sodium chloride solution will be higher

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

Correct answer:

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

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, k_b is the boiling point elevation constant, is the number of ions, and is the molality. Sodium chloride, or NaCl , will produce two ions in solution whereas calcium chloride, or 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.

How does the boiling point of a solution change after adding potassium bromide? The solution has a density of $1.5\frac{g}{mL}$ and $k_b = 0.512\frac{^oC}{m}$ .

Possible Answers:

Increases by 0.741^oC

Decreases by 0.741^oC

Increases by 1.3^oC

Decreases by 1.3^oC

Correct answer:

Increases by 0.741^oC

Explanation:

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.

$\Delta T = k_bim$

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

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

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

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

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

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 $119\frac{g}{mol}$ (this can be calculated by obtaining values from the periodic table). We have $1\:mole$ of solute; therefore, the mass of solute is

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

This means that the mass of solvent is equal to:

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

Molality of solution is

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

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

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

Therefore, boiling point increases by 0.741^oC .

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

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

Which of the following will result in the least freezing point depression when added to $\small 500mL$ of water?

Correct answer:

$10g\ \text{of}\ Ba(OH)_2$

Explanation:

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

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

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.

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

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

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

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

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

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

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

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

$K_2O:\ i=3$

$NaCl:\ i=2$

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

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

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

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

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

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

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

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

What is the freezing point of a $\small 0.2m$ solution of glucose in benzene?

Correct answer:

4.52^oC

Explanation:

First, calculate the freezing point depression with the equation:

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

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.

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

$\Delta T_f=0.98^oC$

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

5.5^oC-0.98^oC=4.52^oC

What is the freezing point of a 3m solution of K₂SO₄ in water? (k_f = 1.9^oC/mol)

Correct answer:

$-17.1^{\circ}C$

Explanation:

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

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.

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

Five solutions are made, each comprised of 1L of water and 1mol of C₆H₁₂O₆, NaCl, H₂SO₄, Al(NO₃)₃, or CaCl₂.

Which solution will show the smallest amount of freezing point depression?

Possible Answers:

H₂SO₄

NaCl

Al(NO₃)₃

C₆H₁₂O₆

CaCl₂

Explanation:

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.

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Find the Boiling Point of the Following Solutions 2.00 M

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Akhurst Robjecia

Find the Boiling Point of the Following Solutions 2.00 M

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