The density of acetonitrile (CH3CN) is 0.786g /mL and the density of methanol (CH3OH) is 0.791g/mL. A solution is made by dissolving 22.5 mL of CH3OH in 98.7 mL of CH3CN . (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of CH3OH in the solution?
98.7 mL CH3CN x 0.786 g/mL = about 77.6g
mols CH3CN = 77.6/41 = about 1.9
22.5 mL CH3OH x 0.791 = estimated about 18 g
mols CH3OH = about 18/32 = about 0.56
a. XCH3OH = mols CH3OH/total mols
b. There is no molality of the solution. Since you're asking for molarity CH3OH in part c I'll assume you want molality of CH3OH.
molality = m = mols CH3OH/kg CH3CN
c. M CH3OH = mols CH3OH/L of solution. volume of solution will be 22.5 mL + 98.7 mL = 121.2 mL assuming they are additive.
Note that I have estimated al along. You should redo each calculation and use your values.
Post your work if you get stuck.
(a) Ah, the mole fraction! Always the life of the party. To find the mole fraction of methanol in the solution, we first need to find the moles of methanol and acetonitrile individually.
For methanol:
Moles of CH3OH = (volume of CH3OH in mL) x (density of CH3OH in g/mL) x (1 mole/ molar mass of CH3OH)
Moles of CH3OH = (22.5 mL) x (0.791 g/mL) x (1 mole/ 32.04 g/mol)
For acetonitrile:
Moles of CH3CN = (volume of CH3CN in mL) x (density of CH3CN in g/mL) x (1 mole/ molar mass of CH3CN)
Moles of CH3CN = (98.7 mL) x (0.786 g/mL) x (1 mole/ 41.05 g/mol)
Now we can calculate the mole fraction of methanol using the formula:
Mole fraction of CH3OH = (moles of CH3OH) / (moles of CH3OH + moles of CH3CN)
(b) That molality though! To find the molality of the solution, we need to calculate the moles of solute (CH3OH) and the mass of the solvent (CH3CN).
Moles of CH3OH = (22.5 mL) x (0.791 g/mL) x (1 mole/ 32.04 g/mol)
Mass of CH3CN = (volume of CH3CN in mL) x (density of CH3CN in g/mL)
Mass of CH3CN = (98.7 mL) x (0.786 g/mL)
Now we can calculate the molality using the formula:
Molality = (moles of CH3OH) / (mass of CH3CN in kg)
(c) Ah, the mysterious molarity! Assuming that volumes are additive (which they are for ideal solutions), we can calculate the molarity of CH3OH.
First, we need to convert the mass of CH3OH to moles:
Moles of CH3OH = (volume of CH3OH in mL) x (density of CH3OH in g/mL) x (1 mole/ molar mass of CH3OH)
Now, we can calculate the molarity using the formula:
Molarity (M) = (moles of CH3OH) / (volume of solution in liters)
Now, let the calculations begin!
To find the mole fraction of methanol (CH3OH) in the solution and the molality of the solution, we need to calculate the moles of each component first.
Step 1: Calculate the moles of CH3OH.
The given volume of CH3OH is 22.5 mL, and its density is 0.791 g/mL.
Therefore, the mass of CH3OH is:
Mass of CH3OH = Volume of CH3OH x Density of CH3OH = 22.5 mL x 0.791 g/mL = 17.7985 g
The molar mass of CH3OH is:
Molar mass of CH3OH = (12.01 g/mol x 1) + (1.01 g/mol x 4) + (16.00 g/mol + 1) = 32.04 g/mol
Now, calculate the moles of CH3OH:
Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH = 17.7985 g / 32.04 g/mol = 0.5557 mol
Step 2: Calculate the moles of CH3CN.
The given volume of CH3CN is 98.7 mL, and its density is 0.786 g/mL.
Therefore, the mass of CH3CN is:
Mass of CH3CN = Volume of CH3CN x Density of CH3CN = 98.7 mL x 0.786 g/mL = 77.5642 g
The molar mass of CH3CN is:
Molar mass of CH3CN = (12.01 g/mol x 1) + (1.01 g/mol x 1) + (12.01 g/mol x 1) + (14.01 g/mol x 1) = 41.05 g/mol
Now, calculate the moles of CH3CN:
Moles of CH3CN = Mass of CH3CN / Molar mass of CH3CN = 77.5642 g / 41.05 g/mol = 1.8900 mol
(a) Mole fraction of methanol (CH3OH):
Mole fraction of CH3OH = Moles of CH3OH / (Moles of CH3OH + Moles of CH3CN)
Mole fraction of CH3OH = 0.5557 mol / (0.5557 mol + 1.8900 mol) = 0.227
(b) Molality of the solution:
Molality = Moles of solute / Mass of solvent in kg
The mass of solvent (CH3CN) = Volume of CH3CN x Density of CH3CN = 98.7 mL x 0.786 g/mL = 77.5642 g
Mass of solvent in kg = 77.5642 g / 1000 g/kg = 0.0776 kg
Molality = Moles of CH3OH / Mass of CH3CN in kg
Molality = 0.5557 mol / 0.0776 kg = 7.156 mol/kg
(c) Assuming volumes are additive, we can find the molarity of CH3OH in the solution.
Molarity = Moles of CH3OH / Volume of solution in L
The volume of the solution = Volume of CH3OH + Volume of CH3CN = 22.5 mL + 98.7 mL = 121.2 mL
Volume of solution in L = 121.2 mL / 1000 mL/L = 0.1212 L
Molarity = Moles of CH3OH / Volume of solution in L
Molarity = 0.5557 mol / 0.1212 L = 4.584 M
Therefore,
(a) The mole fraction of CH3OH in the solution is 0.227.
(b) The molality of the solution is 7.156 mol/kg.
(c) The molarity of CH3OH in the solution is 4.584 M.
To solve these questions, we need to use the formulas for mole fraction, molality, and molarity. I will explain each step of the calculations.
(a) Mole fraction (χ) is the ratio of the number of moles of one component to the total number of moles in the solution. To find the mole fraction of methanol (CH3OH), we need to calculate the moles of methanol (nCH3OH) and acetonitrile (nCH3CN).
1) Calculate the mass of CH3OH:
Mass of CH3OH = density * volume = 0.791 g/mL * 22.5 mL = 17.79 g
2) Convert the mass of CH3OH to moles:
Molar mass of CH3OH = 12.01 g/mol + (3 * 1.01 g/mol) + 16.00 g/mol = 32.04 g/mol
nCH3OH = mass / molar mass = 17.79 g / 32.04 g/mol ≈ 0.556 mol
3) Calculate the moles of CH3CN:
Mass of CH3CN = density * volume = 0.786 g/mL * 98.7 mL = 77.50 g
Molar mass of CH3CN = 12.01 g/mol + 3 * 1.01 g/mol + 12.01 g/mol + 14.01 g/mol = 41.05 g/mol
nCH3CN = mass / molar mass = 77.50 g / 41.05 g/mol ≈ 1.888 mol
4) Calculate the mole fraction of CH3OH:
χCH3OH = nCH3OH / (nCH3OH + nCH3CN) = 0.556 mol / (0.556 mol + 1.888 mol) ≈ 0.227
Therefore, the mole fraction of methanol (CH3OH) in the solution is approximately 0.227.
(b) Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, we will assume that the volumes of CH3OH and CH3CN are additive, so we can consider the total volume as 22.5 mL + 98.7 mL = 121.2 mL = 0.1212 L.
1) Convert the mass of CH3OH to kilograms:
Mass of CH3OH = density * volume = 0.791 g/mL * 22.5 mL = 17.79 g
Mass of CH3OH in kg = 17.79 g / 1000 = 0.01779 kg
2) Calculate the molality:
m = nCH3OH / mass of solvent in kg = 0.556 mol / 0.01779 kg ≈ 31.21 mol/kg
Therefore, the molality of the solution is approximately 31.21 mol/kg.
(c) Molarity (M) is defined as the number of moles of solute per liter of solution. Since we already know the mole fraction of CH3OH, we can use this information to calculate the molarity.
1) Calculate the moles of CH3OH:
nCH3OH = χCH3OH * (nCH3OH + nCH3CN) = 0.227 * (0.556 mol + 1.888 mol) ≈ 0.556 mol
2) Calculate the volume of the solution:
Volume of the solution = volume of CH3OH + volume of CH3CN = 22.5 mL + 98.7 mL = 121.2 mL = 0.1212 L
3) Calculate the molarity:
M = nCH3OH / volume of solution in liters = 0.556 mol / 0.1212 L ≈ 4.588 M
Therefore, the molarity of CH3OH in the solution is approximately 4.588 M.