뒤로General Chemistry: Solutions and Colligative Properties – Step-by-Step Study Guidance
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Q1. 7.20 grams of C2H6O2 is added to 3546 grams of water. Calculate the mole fraction for both C2H6O2 and water.
Background
Topic: Solution Concentration – Mole Fraction
This question tests your understanding of how to calculate mole fractions, which is a way to express the concentration of a component in a mixture.
Key Terms and Formulas
Mole fraction (Χ): The ratio of the moles of one component to the total moles in the solution.
Formula:
Step-by-Step Guidance
Calculate the number of moles of C2H6O2 using its molar mass (find the molar mass by adding up the atomic masses: C, H, and O).
Calculate the number of moles of water (H2O) using its molar mass (18.02 g/mol).
Add the moles of C2H6O2 and water to get the total moles in the solution.
Set up the mole fraction formula for each component: and .
Try solving on your own before revealing the answer!
Final Answer:
Moles of C2H6O2: mol
Moles of H2O: mol
Total moles = 0.116 + 197 = 197.116 mol
Mole fraction of C2H6O2:
Mole fraction of H2O:
The mole fraction of C2H6O2 is very small compared to water, as expected for a dilute solution.
Q2. 25.0 grams of KMnO4(s) (158.034 g/mol) is dissolved in water. The total mass of the solution is 250.0 grams. What is the molality of the solution?
Background
Topic: Solution Concentration – Molality
This question tests your ability to calculate molality, which is the number of moles of solute per kilogram of solvent.
Key Terms and Formulas
Molality (m):
Solute: KMnO4
Solvent: Water (mass of solvent = total mass - mass of solute)
Step-by-Step Guidance
Calculate the moles of KMnO4 using its molar mass.
Find the mass of the solvent (water) by subtracting the mass of KMnO4 from the total mass of the solution.
Convert the mass of the solvent from grams to kilograms.
Set up the molality formula using the moles of solute and kilograms of solvent.
Try solving on your own before revealing the answer!
Final Answer:
Moles of KMnO4: mol
Mass of solvent: g = kg
Molality: mol/kg
The solution has a molality of 0.702 mol/kg.
Q3. An acid solution is 25.0% by mass HNO3(aq) (63.013 g/mol). The density is 1.344 g/mL. What is the molarity of the solution? Hint: Assume 100 g of solution.
Background
Topic: Solution Concentration – Molarity
This question tests your ability to convert between percent by mass, density, and molarity (moles of solute per liter of solution).
Key Terms and Formulas
Molarity (M):
Percent by mass:
Density:
Step-by-Step Guidance
Assume 100 g of solution (so 25.0 g HNO3 and 75.0 g water).
Calculate the moles of HNO3 using its molar mass.
Use the density to find the volume of 100 g of solution in mL, then convert to liters.
Set up the molarity formula using the moles of HNO3 and the volume in liters.
Try solving on your own before revealing the answer!
Final Answer:
Moles of HNO3: mol
Volume of solution: mL = L
Molarity: M
The molarity of the solution is 5.34 M.
Q4. What happens to the freezing point of a solution as you add a solute? What happens to its boiling point?
Background
Topic: Colligative Properties – Freezing Point Depression and Boiling Point Elevation
This question tests your understanding of how adding a solute affects the physical properties of a solvent.
Key Terms and Formulas
Freezing point depression:
Boiling point elevation:
i: van 't Hoff factor (number of particles the solute splits into)
Step-by-Step Guidance
Recall that adding a nonvolatile solute to a solvent lowers the freezing point and raises the boiling point.
Use the formulas above to explain how the magnitude of the change depends on the concentration and the van 't Hoff factor.
Think about why these changes occur in terms of molecular interactions and vapor pressure.
Try answering in your own words before revealing the answer!
Final Answer:
Adding a solute to a solvent lowers the freezing point (freezing point depression) and raises the boiling point (boiling point elevation) of the solution. This is because the solute particles disrupt the formation of the solid phase and require a higher temperature for boiling.
Q5. A liquid solution contains 471 g of acetic acid, HC2H3O2, and 117 g of water, H2O. What is the freezing point of this mixture? H2O = 18.02 g/mol, HC2H3O2 = 60.05 g/mol. Hint: Which compound is actually the solvent?
Background
Topic: Colligative Properties – Freezing Point Depression
This question tests your ability to determine the freezing point of a solution using the freezing point depression formula, and to identify the solvent in a mixture.
Key Terms and Formulas
Freezing point depression:
Freezing point of pure solvent:
Freezing point of solution:
Molality (m):
Step-by-Step Guidance
Determine which component is the solvent (the one in greater amount by moles).
Calculate the moles of solute and solvent using their molar masses.
Calculate the molality of the solution (moles of solute per kg of solvent).
Use the appropriate and values for the solvent to set up the freezing point depression formula.
Try solving on your own before revealing the answer!
Final Answer:
Water is the solvent (more moles than acetic acid).
Moles of acetic acid: mol
Mass of water: 117 g = 0.117 kg
Molality: mol/kg
Freezing point depression: °C
Freezing point: °C
The freezing point of the mixture is -124.6 °C (using water as the solvent).
Q6. What happens to the vapor pressure of a pure liquid as you increase the temperature? Does the vapor pressure change linearly with temperature? What does the liquid do once its vapor pressure is the same as the pressure surrounding the vessel containing the liquid?
Background
Topic: Vapor Pressure and Boiling
This question tests your understanding of how vapor pressure changes with temperature and the concept of boiling.
Key Terms and Formulas
Vapor pressure: The pressure exerted by a vapor in equilibrium with its liquid at a given temperature.
Clausius-Clapeyron equation:
Step-by-Step Guidance
Recall that vapor pressure increases as temperature increases.
Understand that the relationship is not linear; it is exponential, as described by the Clausius-Clapeyron equation.
When the vapor pressure equals the external pressure, the liquid boils.
Try answering in your own words before revealing the answer!
Final Answer:
The vapor pressure of a pure liquid increases exponentially with temperature (not linearly). When the vapor pressure equals the surrounding pressure, the liquid boils.
Q7. Use information from the data table below to calculate the enthalpy of vaporization, ΔHvap, for liquid ammonia, NH3. R = 8.314 J/(mol ∙ K)
Temperature (°C): -60.8, -46.3 Pressure (atm): 0.197, 0.395
Background
Topic: Phase Changes – Enthalpy of Vaporization
This question tests your ability to use the Clausius-Clapeyron equation to calculate the enthalpy of vaporization from vapor pressure and temperature data.
Key Terms and Formulas
Clausius-Clapeyron equation:
Temperatures must be in Kelvin.
Step-by-Step Guidance
Convert both temperatures from Celsius to Kelvin by adding 273.15.
Calculate using the given pressures.
Calculate using the Kelvin temperatures.
Rearrange the Clausius-Clapeyron equation to solve for .
Try solving on your own before revealing the answer!
Final Answer:
Convert temperatures: K, K
K
J/mol = 19.1 kJ/mol
The enthalpy of vaporization for ammonia is 19.1 kJ/mol.
Q8. Which of the following aqueous solutions would have the lowest freezing point? Which would have the highest boiling point? Kf H2O = 1.858 °C/molal, Kb H2O = 0.512 °C/molal. a. 0.15 mol/kg C6H12O6 b. 0.15 mol/kg NaCl c. 0.15 mol/kg CaCl2
Background
Topic: Colligative Properties – Freezing Point Depression and Boiling Point Elevation
This question tests your understanding of how the van 't Hoff factor (i) affects colligative properties for different solutes.
Key Terms and Formulas
Freezing point depression:
Boiling point elevation:
i (van 't Hoff factor): Number of particles the solute produces in solution
Step-by-Step Guidance
Determine the van 't Hoff factor (i) for each solute: C6H12O6 (nonelectrolyte, i=1), NaCl (i=2), CaCl2 (i=3).
Calculate for each solution to compare the effect on freezing and boiling points.
The solution with the largest will have the greatest freezing point depression (lowest freezing point) and highest boiling point elevation.
Try answering before revealing the answer!
Final Answer:
CaCl2 (i=3) gives the largest value, so it has the lowest freezing point and highest boiling point. C6H12O6 (i=1) has the smallest effect.
Q9. Calculate the vapor pressure of a solution that contains 250.0 g of C6H6 (78.10 g/mol) and 50.1 g of lauric acid, C12H24O2 (200.3 g/mol). The vapor pressure of C6H6 at 20.0 °C is 74.6 mmHg.
Background
Topic: Colligative Properties – Raoult’s Law
This question tests your ability to use Raoult’s Law to calculate the vapor pressure of a solution containing a nonvolatile solute.
Key Terms and Formulas
Raoult’s Law:
: Vapor pressure of pure solvent (C6H6)
: Mole fraction of solvent
Step-by-Step Guidance
Calculate the moles of C6H6 and lauric acid using their molar masses.
Find the total moles in the solution.
Calculate the mole fraction of C6H6.
Use Raoult’s Law to set up the calculation for the vapor pressure of the solution.
Try solving on your own before revealing the answer!
Final Answer:
Moles C6H6: mol
Moles lauric acid: mol
Total moles = 3.20 + 0.250 = 3.45 mol
Mole fraction C6H6:
Vapor pressure: mmHg
The vapor pressure of the solution is 69.2 mmHg.
Q10. A saline solution is prepared by dissolving 9.00 g of NaCl (58.44 g/mol) in enough water to make 1.00 L of solution. What is the osmotic pressure of this solution at 37.0 °C?
Background
Topic: Colligative Properties – Osmotic Pressure
This question tests your ability to calculate osmotic pressure using the van 't Hoff equation.
Key Terms and Formulas
Osmotic pressure:
i: van 't Hoff factor for NaCl (i=2)
M: Molarity (moles of solute per liter of solution)
R: Gas constant (0.08206 L·atm/mol·K)
T: Temperature in Kelvin
Step-by-Step Guidance
Calculate the moles of NaCl using its molar mass.
Find the molarity by dividing moles by the volume of solution in liters.
Convert the temperature from Celsius to Kelvin.
Set up the osmotic pressure formula using the calculated values and the appropriate gas constant.
Try solving on your own before revealing the answer!
Final Answer:
Moles NaCl: mol
Molarity: M
Temperature: K
Osmotic pressure: atm
The osmotic pressure of the solution is 7.82 atm.