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17.11: Worksheets for Chapter 17

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Concentration by Percent Mass Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

The definition of percent mass concentration is the ratio of the mass of solute divided by the total mass of the solution and multiplied by 100 to convert to a percentage.

$percent \ by \ mass = \frac {mass \ of \ solute} {mass \ of \ solution} \times \ 100$

Example: What is the percent concentration by mass of a solution formed by dissolving $100. \ grams$ of ethanol, $C_2H_5OH$, in $100. \ grams$ of water?

Solution: percent by mass = $\frac {mass \ of \ solute} {mass \ of \ solution} \times 100 = \frac {100. \ g} {200. \ g} \times 100 = 50.0 \%$

Example: If the density of a $10.0\%$ by mass $KNO_3$ solution in water is $1.19 \ g/mL$, how many grams of $KNO_3$ are present in $100. \ mL$ of the solution?

Solution: We can multiply the volume times the density to the mass of the $100. \ mL$ of solution and then take $10.0\%$ of the mass of the solution to get the mass of the potassium nitrate.

$grams \ of \ solution &= (100. \ mL)(1.19 \ g/mL) = 119 \ grams\\grams \ of \ KNO_3 &= (0.10)(119 \ grams) = 11.9 \ grams$

Exercises

1. If $30.0 \ grams$ of $AgNO_3$ are dissolved in $275 \ grams$ of water, what is the concentration of the silver nitrate by mass percent?
2. How many grams of $MgF_2$ are present in $100.0 \ g$ of a $20.0\%$ $MgF_2$ in water solution?
3. How many grams of water are present in the solution in question #2?
4. The density of a $30.0\%$ by mass solution of $NaOH$ in water is $1.33 \ g/mL$. How many grams of $NaOH$ are required to prepare $500. \ mL$ of this solution?
5. The density of pure water is $1.00 \ g/mL$. What is the concentration gy percent mass of a solution prepared by dissolving $85.0 \ grams$ of $NaOH$ in $750. \ mL$ of water?
6. A solution is prepared by dissolving $66.0 \ grams$ of acetone, $C_3H_6O$, in $146.0 \ grams$ of water. The density of the solution is $0.926 \ g/mL$. What is the percent concentration of acetone by mass?
7. A $35.4\%$ solution of $H_3PO_4$ in water has a density of $1.20 \ g/mL$. How many grams of phosphoric acid are present in $300. \ mL$ of this solution?

Mole Fraction and Molality Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

Mole Fraction

The definition of mole fraction is the ratio of the moles of solute divided by the total moles of the solution.

$mole \ fraction = \frac {moles \ of \ solute} {moles \ of \ solution}$

Example: What is the mole fraction of ethanol in a solution prepared by dissolving $100. \ g$ of ethanol, $C_2H_5OH$, in $100. \ g$ of water?

Solution:

$moles \ ethanol &= \frac {100. \ g} {46.0 \ g/mol} = 2.17 \ moles\\moles \ water &= \frac {100. \ g} {18.0 \ g/mol} = 5.56 \ moles\\mole \ fraction \ of \ ethanol &= \frac {2.17 \ mols} {7.73 \ mols} = 0.281$

Molality

The definition of molality is the ratio of the moles of solute divided by the kilograms of solvent.

$molality = \frac {moles \ of \ solute} {kilograms \ of \ solvent}$

Example: What is the molality of a solution prepared by dissolving $100. \ g$ of ethanol, $C_2H_5OH$, in $100. \ g$ of water?

$moles \ ethanol &= \frac {100. \ g} {46.0 \ g/mol} = 2.17 \ moles\\molality \ of \ ethanol & = \frac {2.17 \ mols} {0.100 \ kg} = 21.7 \ m$

Example: A $35.4\%$ solution of $H_3PO_4$ in water has a density of $1.20 \ g/mL$. What is the mole fraction of $H_3PO_4$ in this solution and what is the molality?

Solution: We can choose a sample volume of this solution and get the mass of it by multiplying the volume times the density. Suppose we choose a $1.00 \ L$ sample.

$\text{mass of solution} &= (1000. \ mL)(1.20 \ g/mL) = 1200. \ grams\\\text {mass of} \ H_3PO_4 \ \text{in the solution} &= (0.354)(1200. \ grams) = 425 \ grams\\\text{mass of} \ H_2O &= 1200. \ grams - 425 \ grams = 775 \ grams\\moles \ H_3PO_4 &= \frac {425 \ g} {98.0 \ g/mol} = 4.34 \ moles\\moles \ H_2O &= \frac {775 \ g} {18.0 \ g/mol} = 43.1 \ moles\\mole \ \text{fraction of} \ H_3PO_4 &= \frac {4.34 \ mol} {47.4 \ mol} = 0.0916 \\molality & = \frac {4.34 \ mol} {0.775 \ kg} = 5.60 \ m$

Exercises

1. What is the mole fraction of $MgF_2$ in a solution that has $20.0 \ g$ of $MgF_2$ dissolved in $80.0 \ grams$ of water?
2. What is the molality of the solution in question 1?
3. The density of a $30.0\%$ by mass solution of $NaOH$ in water is $1.33 \ g/mL$. What is the mole fraction of $NaOH$ in this solution?
4. What is the molality of the solution in problem 3?
5. What is the molality of a solution prepared by dissolving $4.00 \ g$ of $NaCl$ in $100. \ g$ of water?
6. How many grams of beryllium chloride would you need to add to $125 \ g$ of water to make a $0.500 \ m$ solution?
7. What would be the mole fraction of $BeCl_2$ in the solution in problem 6?
8. A solution is prepared by dissolving $66.0 \ g$ of acetone, $C_3H_6O$, in $146.0 \ g$ of water. The density of the solution is $0.926 \ g/mL$. What is the molality of this solution?
9. What is the mole fraction of acetone in the solution in problem 8?

Molarity Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

The definition of molarity is the ratio of the mols of solute divided by the volume of the solution.

$molarity = \frac {moles \ of \ solute}{liters \ of \ solution}$

Example: What is the molarity of a solution prepared by dissolving $60.0 \ grams$ of $NaOH$ in sufficient water to produce $2.00 \ liters$ of solution?

Solution:

$moles \ NaOH & = \frac {60.0 \ g} {40.0 \ g/mol} = 1.50 \ moles\\molarity & = \frac {1.50 \ mol} {2.00 \ L} = 0.750 \ M$

Example: What volume of $0.750 \ M \ NaOH$ solution will contain $10.0 \ gram$ of $NaOH$?

$moles \ NaOH & = \frac {10.0 \ g} {40.0 \ g/mol} = 0.250 \ moles\\volume = \frac {mol}{M} & = \frac {0.250 \ mol} {0.750 \ mol/L} = 0.333 \ L$

Exercises

1. What is the molarity of a solution in which $4.50 \ g$ of $NaNO_3$ is dissolved in $265 \ mL$ of solution?
2. How many grams of ammonia, $NH_3$ are present in $5.0 \ L$ of $0.100 \ M$ solution?
3. How many milliliters of $0.200 \ M \ NaOH$ solution is necessary to contain $6.00 \ grams$ of $NaOH$?
4. How many liters of $0.500 \ M \ CaF_2$ solution is required to contain $78.0 \ g$ of $CaF_2$?
5. What mass of ammonium phosphate is needed to make $100. \ mL$ of $0.500 \ M \ (NH_4)_3PO_4$ solution?
6. What is the molarity of a solution prepared by dissolving $198 \ g$ of $BaBr_2$ in $2.00 \ liters$ of solution?
7. How many grams of glycerine, $C_3H_8O_3$, are needed to make $100. \ mL$ of $2.60 \ M$ solution?
8. A test tube contains $10.0 \ mL$ of $3.00 \ M \ CaCO_3$ solution. How many grams of calcium carbonate are in the tube?

Dilution Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

The process of dilution involves increasing the amount of solvent in a solution without changing the amount of solute. For example, you could dilute $50. \ mL$ of $0.250 \ M$ $HCl$ solution by placing the solution in a $100. \ mL$ graduated cylinder and adding water until the solution reached the $100. \ mL$ line in the graduate. The original solution contained $0.0125 \ moles$ of $HCl$ before it was diluted and therefore, it also contains $0.0125 \ moles$ of $HCl$ after the dilution. In the process of dilution, the amount of solute never changes. The amount of solvent, the total volume of the solution, and the concentration change but the amount of solute remains the same.

For a solution whose concentration is expressed in molarity, the moles of solute can be calculated by multiplying the volume in liters times the molarity.

$moles \ solute = (molarity)(liters)$

For the moles of solute in the original solution, $moles_{initial} = molarity_{initial} \times liters_{initial}$ or $mols_i = M_i \times V_i$. After the solution has been diluted, the moles in the final solution can be calculated with $mols_f = M_f \times V_f$. Since the mols do not change during dilution,

$mols_i = mols_f \quad \text{and} \quad M_i \times V_i = M_f \times V_f.$

In the dilution problems you will be given, for the most part, three of the four variables or ways to find three of the four variables and you will asked to calculate the fourth variable.

Example: How many milliliters of $6.00 \ M \ NaOH$ solution are necessary to prepare $300. \ mL$ of $1.20 \ M \ NaOH$ solution?

Solution:

$(M_i)(V_i) &= (M_f)(V_f)\\V_i &= \frac {(M_f)(V_f)} {(M_i)} = \frac {(1.20 \ M)(0.300 \ L)} {(6.00 \ M)} = 0.0600 \ L = 60.0 \ mL$

Exercises

1. $200. \ mL$ of $3.00 \ M \ NaCl$ solution is diluted to a final volume of $500. \ mL$. What is the concentration of the final solution?
2. $100. \ mL$ of concentrated hydrochloric acid was diluted to $1.20 \ liters$ of $1.00 \ M$ solution. What was the concentration of the original concentrated solution?
3. What volume of $6.00 \ M \ NaOH$ is needed to prepare $250. \ mL$ of $0.600 \ M \ NaOH$?
4. If $25.0 \ mL$ of $16.0 \ M \ HNO_3$ is diluted to $500. \ mL$, what is the final concentration?
5. To what volume must you dilute $10.0 \ mL$ of $6.00 \ M \ H_2SO_4$ to produce a solution that is $1.00 \ M \ H_2SO_4$?
6. Solution A is $5.00 \ mL$ of $12.0 \ M \ HCl$. Solution B is prepared by diluting solution A to a new volume of $100. \ mL$. Solution C is produced by taking $5.00 \ mL$ of solution B and diluting it to $100. \ mL$. What is the molarity of solution C?

Colligative Properties: Solution Vapor Pressure Worksheet

Colligative properties are those properties of a solution that depend on the number of particles of solute present in the solution, and not on the chemistry nor the mass of the particles. That is, the chemical behavior and the molar masses of urea, $(NH_2)_2CO$, and glucose, $C_6H_{12}O_6$, are very different, but the colligative properties of a $1.0 \ M$ solution of urea will be exactly the same as the colligative properties of a $1.0 \ M$ solution of glucose.

The colligative properties of solutions include vapor pressure lowering, boiling point elevation, freezing point depression, and changes in osmotic pressure. The changes in these properties are dependent entirely on the concentration of particles of solute in the solution. It must be noted that ionic solutes dissociate when dissolved in water and therefore, add more particles to the solution than a substance that does not dissociate in water.

Vapor Pressure Lowering

The vapor pressure of a solution can be calculated from the individual vapor pressures of the components (solute and solvent) and the mole fractions of each component. Raoult's Law is an expression of the relationship.

$& \text{Vapor Pressure}_{\text{solution}} = (X_{\text{mol fraction solvent}})(\text{Vapor Pressure}_{\text{solvent}})\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad +\\& \qquad \qquad \qquad \qquad \qquad \ \ (X_{\text{mol fraction solute}})(\text{Vapor Pressure}_{\text{solute}})$

Example: What is the vapor pressure, at $25^ \circ C$, of a solution produced by dissolving $50.0$ of acetone, $C_3H_6O$, in $50.0 \ grams$ of water? The vapor pressure of pure acetone at $25^ \circ C$ is $230. \ mm$ of $Hg$ and the vapor pressure of pure water at $25^ \circ C$ is $23.7 \ mm$ of $Hg$.

Solution: $50.0 \ g$ of acetone is $0.86 \ moles$ and $50.0 \ g$ of water is $2.78 \ moles.$

Therefore, the mole fractions in this solution are $0.236$ acetone and $0.764$ water.

$VP_{\text{SOLUTION}} & = (0.764)(23.7 \ mm \ of \ Hg) + (0.236)(230. \ mm \ of \ Hg)\\& = 18.1 \ mm \ of \ Hg + 54.3 \ mm \ of \ Hg \\& = 72.4 \ mm \ of \ Hg$

In this case, the vapor pressure of the solution is higher than the vapor pressure of the solvent. That is due to the fact that acetone is a volatile (weak intermolecular forces of attraction) and therefore, evaporates readily. When we refer to vapor pressure lowering, we are referring to solutions in which the solute is non-volatile. When the solute is a solid, it can be generally be assumed that the solute is non-volatile.

Suppose we are making a solution of glucose in water. Glucose is a non-volatile, solid solute whose vapor pressure at room conditions is so small that it is negligible compared to the vapor pressure of water. When we substitute the values for a glucose solution into Raoult's Law, the second term (the one for the solute) is essentially zero because the vapor pressure of the pure solute is essentially zero.

$& \text{Vapor Pressure}_{\text{Solution}} = (X_{\text{Mol fraction solvent}})(\text{Vapor Pressure}_{\text{Solvent}})\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad +\\& \qquad \qquad \qquad \qquad \qquad \ \ (X_{\text{Mol fraction solute}})(\text{Vapor Pressure}_{\text{Solute}})$

If the second term in this equation, $(X_{\text{Mol fraction solute}})(\text{Vapor Pressure}_{\text{Solute}})$, becomes zero, then for a solution with a non-volatile solute, Raoult's Law becomes:

$\text{Vapor Pressure}_{\text{Solution}} = (X_\text{Mol fraction solvent})(\text{Vapor Pressure}_{\text{Solvent}})$

This is Raoult's Law for solutions whose solute is a non-volatile.

$\text{VP}_{\text{Solution}} = (X_{\text{Solvent}})(\text{VP}_{\text{Solvent}})$

Example: What is the vapor pressure, at $25^o$C, of a solution produced by dissolving $50.0$ of glucose, $25^ \circ C$, in $50.0 \ grams$ of water? Glucose is non-volatile and the vapor pressure of pure water at $25^ \circ C$ is $23.7 \ mm$ of $Hg$.

Solution: $50.0 \ g$ of water is $2.78 \ moles$ and $50.0 \ g$ of glucose is $0.278 \ moles.$

Therefore, the mole fraction of water in this solution is $0.909$. We do not need to calculate the mole fraction of glucose because it isn't needed in Raoult's Law for non-volatile solutes.

$\text{VP}_{\text{Solution}} = (X_{\text{Solvent}})(\text{VP}_{\text{Solvent}} = (0.909)(23.7 \ mm \ of \ Hg) = 21.5 \ mm \ of \ Hg$

In this case, and in all cases of non-volatile solutes, the vapor pressure of the solution is less than the vapor pressure of the pure solvent.

Exercises

1. If $25.0 \ grams$ of sodium chloride is added to $500. \ grams$ of water at $25^ \circ C$, what will be the vapor pressure of the resulting solution in kPa? The vapor pressure of pure water at $25^ \circ C$ is $3.17 \ kPa.$
2. $125 \ g$ of the non-volatile solute glucose, $C_6H_{12}O_6$, is dissolved in $125 \ g$ of water at $25^ \circ C$. IF the vapor pressure of water at $25^ \circ C$ is $23.7 \ Torr$, what is the vapor pressure of the solution?
3. Glycerin, $C_3H_8O_3$, is a non-volatile, non-electrolyte solute. If $53.6 \ g$ of glycerin is dissolved in $133.7 \ g$ of ethanol at $40.^ \circ C, C_2H_5OH$, what is the vapor pressure of the solution? The vapor pressure of pure ethanol is $113 \ Torr$ at $40.^ \circ C.$
4. The vapor pressure of hexane, $C_6H_{14}$, at $60.0^ \circ C$ is $573 \ Torr$. The vapor pressure of benzene at the same temperature is $391 \ Torr$. What will be the vapor pressure of a solution of $58.9 \ g$ of hexane with $44.0 \ g$ of benzene?

Colligative Properties: B.P. Elevation and M.P. Depression Worksheet

When a non-volatile, solid solute is added to a solvent, the boiling point of the solution will be higher than the boiling point of the solvent, and the melting point of the solution will be lower than the melting point of the solvent. The size of the boiling point elevation and the melting point depression are colligative properties, that is, they are dependent not on the chemistry of the solute but only on the number of solute particles present in the solution.

The formula used to calculate boiling point elevation is $\Delta T_b = imK_b$, where $\Delta T_b$ is the increase in the boiling point, $m$ is the molality of the solute, $K_b$ is the boiling point elevation constant, and $i$ is the van't Hoff factor.

The boiling point elevation constant, $K_b$, is an experimentally determined constant for the solvent. Each solvent will have its own $K_b$ and these values are determined in the laboratory and listed in reference tables. For example, the boiling point elevation constant for water is $0.512^ \circ C/m$. As the molality of the solution increases, the boiling point of the solution increases by $0.512^ \circ C$ for each increase of $1.00$ in the molality.

The van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of the molecules dissolved. For most non-electrolytes dissolved in water, the van't Hoff factor is essentially 1. For most ionic compounds dissolved in water, the van't Hoff factor is equal to the number of discrete ions in a formula unit of the substance. For example, a glucose solution that is $1.00 \ molal$ will have a particle concentration that is also $1.00 \ molal$ because glucose molecules do not dissociate. A $1.00 \ molal$ sodium chloride solution, on the other hand, since it dissociates into two ions will have a particle molality of $2.00 \ m$. The van't Hoff factor, $i$, is the number of ions that the molecule will dissociate into when dissolved. Sometimes, in concentrated solutions, an ionic substance does not dissociate $100\%$ and therefore, the value of $i$ will not be exactly equal to the apparent number of ions produced. In such cases, the value of $i$ must also be determined experimentally. If you are not given an actual value for $i$ in the problem, assume that $i$ is the number of ions apparently produced per molecule. This is true in most dilute solutions.

The formula used to calculate melting point depression is $\Delta T_f = imK_f$, where $\Delta T_f$ is the decrease in the melting point, $m$ is the molality of the solute, $K_f$ is the melting point depression constant, and $i$ is the van't Hoff factor.

The melting point depression constant, $K_f$, is an experimentally determined constant for the solvent. Each solvent will have its own $K_f$ and these values are determined in the laboratory and listed in reference tables. For example, the freezing point depression constant for water is $1.86^ \circ C/m$. As the molality of the solution increases, the melting point of the solution decreases by $1.86^ \circ C$ for each increase of $1.00$ in the molality.

Example: What is the boiling point of a $5.00 \ m$ glucose solution in water? Glucose is a non-volatile, non-electrolyte solute. $K_b$ for water = $0.512^ \circ C/m.$

Solution: $\Delta T_b = imK_b = (1)(5.00 \ m)(0.512^ \circ C/m) = 2.56^ \circ C$

Since the boiling point of the pure solvent was $100.00^ \circ C$, the b.p. of the solution is $100.00^ \circ C + 2.56^ \circ C = 102.56^ \circ C$

Example: What is the melting point of a $5.00 \ m \ NaCl$ solution in water? Sodium chloride is a non-volatile solute that dissociates $100\%$ in water. $K_f$ for water = $1.86^ \circ C/m.$

Solution: $\Delta T_f = imK_f = (2)(5.00 \ m)(1.86^ \circ C/m) = 18.6^ \circ C$ (Since $NaCl$ produces two ions in solution, $i = 2.$)

Since the melting point of the pure solvent was $0.00^ \circ C$, the m.p. of the solution is $0.00^ \circ C - 18.6^ \circ C = - 18.6^ \circ C$

Exercises

1. What is the melting point of a solution produced by dissolving $45.0 \ g$ of $NaCl$ in $500. \ g$ of water. $K_f$ for water = $1.86^ \circ C/m$.
2. What is the boiling point of a solution produced by dissolving $45.0 \ g$ of $NaCl$ in $500. \ g$ of water. $K_b$ for water = $0.512^ \circ C/m$.
3. Which solution will have higher boiling point: a solution containing $105 \ g$ of $C_{12}H_{22}O_{11}$ in $500. \ g$ of water or a solution containing $35.0 \ g$ of $NaCl$ in $500. \ g$ of water?
4. When $25.0 \ g$ of an unknown, non-volatile, non-electrolyte is dissolved in $130. \ g$ of water, the boiling point of the solution is $102.5^ \circ C$. What is the molar mass of the unknown?
5. How many grams of $C_2H_6O_2$ (anti-freeze, a non-electrolyte) must be added to $4,000. \ grams$ of water to reduce the melting point to $-40.^ \circ C$?
6. The melting point constant for benzene is $4.90^ \circ C/m$. The normal melting point of benzene is $5.50^ \circ C$. What is the melting point of a solution of $9.30 \ g$ of $C_{12}H_{25}OH$ (a non-electrolyte) in $250. \ g$ of benzene?
7. Assuming $100\%$ dissociation, what is the boiling point of a solution of $200. \ g$ of $AlF_3$ in $500. \ g$ of water?

Reactions Between Ions in Solution Worksheet

CK-12 Foundation Chemistry

Name______________________ Date_________

For the following five reactions (all reactants are in water solution):

• Write and balance the molecular equation indicating the state of each reactant and product.
• Write the total ionic equation.
• Identify the precipitate.
• Identify the spectator ions.
• Write the net ionic equation.

1. iron (III) chloride + sodium hydroxide

Balanced molecular equation _____________

Total ionic equation _____________

Precipitate = _____________ Spectator ions = _____________

Net ionic equation _____________

2. barium chloride + silver nitrate

Balanced molecular equation _____________

Total ionic equation _____________

Precipitate = _____________ Spectator ions = _____________

Net ionic equation _____________

3. magnesium sulfate + potassium phosphate

Balanced molecular equation _____________

Total ionic equation _____________

Precipitate = _____________ Spectator ions = _____________

Net ionic equation _____________

4. copper (II) nitrate + calcium hydroxide

Balanced molecular equation _____________

Total ionic equation _____________

Precipitate = _____________ Spectator ions = _____________

Net ionic equation _____________

5. sodium chromate + strontium nitrate

Balanced molecular equation _____________

Total ionic equation _____________

Precipitate = _____________ Spectator ions = _____________

• The worksheet answer keys are available upon request. Please send an email to teachers-requests@ck12.org to request the worksheet answer keys.

Aug 18, 2012

Aug 13, 2014