The density of acetonitrile (CH3CN)(CH3CN) is 0.786 g/mLg/mL, and the density of methanol (CH3OH)(CH3OH) is 0.791 g/mLg/mL. A solution is made by dissolving 21.5 mLmL CH3OHCH3OH in 98.8 mLmL CH3CNCH3CN. 1. What is the mole fraction of methanol in the solution? 2. What is the molality of the solution? Assuming CH3OHCH3OH is the solute and CH3CNCH3CN is the solvent. 3. Assuming that the volumes are additive, what is the molarity of CH3OHCH3OH in the solution?
The Correct Answer and Explanation is:
To solve the problem step-by-step, let’s go over each part.
1. Mole Fraction of Methanol in the Solution
Mole fraction is the ratio of the number of moles of a substance to the total number of moles in the solution. We can calculate the mole fraction of methanol (CH₃OH) in the solution using the following steps:
Step 1: Find the mass of CH₃OH and CH₃CN
- Mass of CH₃OH (methanol):
The volume of methanol is given as 21.5 mL, and its density is 0.791 g/mL. Mass of CH₃OH=Density of CH₃OH×Volume of CH₃OH\text{Mass of CH₃OH} = \text{Density of CH₃OH} \times \text{Volume of CH₃OH}Mass of CH₃OH=Density of CH₃OH×Volume of CH₃OH Mass of CH₃OH=0.791 g/mL×21.5 mL=17.0 g\text{Mass of CH₃OH} = 0.791 \, \text{g/mL} \times 21.5 \, \text{mL} = 17.0 \, \text{g}Mass of CH₃OH=0.791g/mL×21.5mL=17.0g - Mass of CH₃CN (acetonitrile):
The volume of acetonitrile is 98.8 mL, and its density is 0.786 g/mL. Mass of CH₃CN=0.786 g/mL×98.8 mL=77.7 g\text{Mass of CH₃CN} = 0.786 \, \text{g/mL} \times 98.8 \, \text{mL} = 77.7 \, \text{g}Mass of CH₃CN=0.786g/mL×98.8mL=77.7g
Step 2: Calculate the moles of CH₃OH and CH₃CN
- Moles of CH₃OH:
The molar mass of CH₃OH is 32.04 g/mol. Moles of CH₃OH=Mass of CH₃OHMolar mass of CH₃OH=17.0 g32.04 g/mol=0.531 mol\text{Moles of CH₃OH} = \frac{\text{Mass of CH₃OH}}{\text{Molar mass of CH₃OH}} = \frac{17.0 \, \text{g}}{32.04 \, \text{g/mol}} = 0.531 \, \text{mol}Moles of CH₃OH=Molar mass of CH₃OHMass of CH₃OH=32.04g/mol17.0g=0.531mol - Moles of CH₃CN:
The molar mass of CH₃CN is 41.05 g/mol. Moles of CH₃CN=Mass of CH₃CNMolar mass of CH₃CN=77.7 g41.05 g/mol=1.89 mol\text{Moles of CH₃CN} = \frac{\text{Mass of CH₃CN}}{\text{Molar mass of CH₃CN}} = \frac{77.7 \, \text{g}}{41.05 \, \text{g/mol}} = 1.89 \, \text{mol}Moles of CH₃CN=Molar mass of CH₃CNMass of CH₃CN=41.05g/mol77.7g=1.89mol
Step 3: Calculate the mole fraction of CH₃OH
Mole fraction of CH₃OH=Moles of CH₃OHMoles of CH₃OH+Moles of CH₃CN=0.531 mol0.531 mol+1.89 mol=0.219\text{Mole fraction of CH₃OH} = \frac{\text{Moles of CH₃OH}}{\text{Moles of CH₃OH} + \text{Moles of CH₃CN}} = \frac{0.531 \, \text{mol}}{0.531 \, \text{mol} + 1.89 \, \text{mol}} = 0.219Mole fraction of CH₃OH=Moles of CH₃OH+Moles of CH₃CNMoles of CH₃OH=0.531mol+1.89mol0.531mol=0.219
2. Molality of the Solution
Molality is defined as the number of moles of solute (CH₃OH) per kilogram of solvent (CH₃CN).
Step 1: Find the mass of CH₃CN in kilograms
The mass of CH₃CN is 77.7 g, which is equivalent to:Mass of CH₃CN=77.7 g=0.0777 kg\text{Mass of CH₃CN} = 77.7 \, \text{g} = 0.0777 \, \text{kg}Mass of CH₃CN=77.7g=0.0777kg
Step 2: Calculate the molality
Molality=Moles of CH₃OHMass of CH₃CN (kg)=0.531 mol0.0777 kg=6.83 mol/kg\text{Molality} = \frac{\text{Moles of CH₃OH}}{\text{Mass of CH₃CN (kg)}} = \frac{0.531 \, \text{mol}}{0.0777 \, \text{kg}} = 6.83 \, \text{mol/kg}Molality=Mass of CH₃CN (kg)Moles of CH₃OH=0.0777kg0.531mol=6.83mol/kg
3. Molarity of CH₃OH in the Solution
Molarity is the number of moles of solute (CH₃OH) per liter of solution.
Step 1: Calculate the total volume of the solution
The total volume is the sum of the volumes of CH₃OH and CH₃CN:Total volume=21.5 mL+98.8 mL=120.3 mL=0.1203 L\text{Total volume} = 21.5 \, \text{mL} + 98.8 \, \text{mL} = 120.3 \, \text{mL} = 0.1203 \, \text{L}Total volume=21.5mL+98.8mL=120.3mL=0.1203L
Step 2: Calculate the molarity
Molarity=Moles of CH₃OHTotal volume of solution (L)=0.531 mol0.1203 L=4.42 M\text{Molarity} = \frac{\text{Moles of CH₃OH}}{\text{Total volume of solution (L)}} = \frac{0.531 \, \text{mol}}{0.1203 \, \text{L}} = 4.42 \, \text{M}Molarity=Total volume of solution (L)Moles of CH₃OH=0.1203L0.531mol=4.42M
Summary of Results:
- The mole fraction of methanol (CH₃OH) in the solution is 0.219.
- The molality of the solution is 6.83 mol/kg.
- The molarity of CH₃OH in the solution is 4.42 M.
This solution involves a detailed process of using the given densities, volumes, and molar masses to find the necessary properties, and the assumption that volumes are additive is critical in determining the molarity.
