A solution was made by dissolving 3.75 g of a pure non volatile solute in 95 g of acetone. The boiling point of pure acetone was observed to be 55.95°c, and that of the solution was 56.50°c. If the molar boiling point elevation constant of acetone is 1.71°c kg/mol, what is the approximate molar mass of the solute?
delta T = i*Kb*m
delta T = 56.50 - 55.95 = ?
i = 1
Kb = 1.71
m = molality. Plug delta T, i and Kb as I've listed them and solve for molality.
Then molality = m = mol/kg solvent. Kg solvent = 95 g acetone. Use 0.095 kg. You have kg and you have molality. Solve for mol. Then mols = grams/molar mass. You have mol and grams, solve for molar mass. Post your work if you get stuck.
Given: mass of solute = 3.75 g
mass of solvent = 95 g = 0.095 kg
boiling point of solvent = 55.95°C
boiling point of solution = 56.50°C
Kb = 1.71°C·kg/mol .
Moles of acetone = \frac{95}{58}=1.64 moled
58
95
=1.64moled
Molarity = \frac{58}{1000}×\frac{328.95×329.5}{1.71}= 3676.35=
1000
58
×
1.71
328.95×329.5
=3676.35
To find the approximate molar mass of the solute, we can use the formula for boiling point elevation:
ΔTb = Kb * m
where:
ΔTb = change in boiling point
Kb = molal boiling point elevation constant
m = molality of the solution
First, let's calculate the change in boiling point:
ΔTb = Tb(solution) - Tb(solvent)
= 56.50°C - 55.95°C
= 0.55°C
Next, let's calculate the molality (m) of the solution:
molality (m) = moles of solute / mass of solvent (in kg)
First, we need to convert the mass of the solute to moles:
moles of solute = mass of solute / molar mass of solute
Given that the mass of the solute is 3.75 g, we can calculate the moles of solute using an approximate molar mass (M):
moles of solute = 3.75 g / M
Now, let's calculate the mass of the solvent (acetone) in kg:
mass of solvent = 95 g = 0.095 kg
Now we can calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
= (3.75 g / M) / 0.095 kg
= 39.47 / M
Finally, we can substitute the values into the boiling point elevation formula to solve for the molar mass of the solute (M):
0.55°C = 1.71°C kg/mol * (39.47 / M)
Simplifying the equation:
0.55 = 1.71 * (39.47 / M)
Rearranging the equation:
39.47 / M = 0.55 / 1.71
Solving for M:
M = (39.47 * 1.71) / 0.55
= 122.35 g/mol
Therefore, the approximate molar mass of the solute is approximately 122.35 g/mol.
To find the molar mass of the solute, we need to use the formula for boiling point elevation:
ΔTb = Kbm
Where:
ΔTb is the boiling point elevation
Kb is the molar boiling point elevation constant of the solvent (acetone)
m is the molality of the solution
First, let's calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.
Step 1: Calculate the number of moles of solute.
To find the number of moles, we use the formula:
moles = mass / molar mass
Given:
mass of solute = 3.75 g
molar mass of solute = ?
We need to find the molar mass of the solute.
Step 2: Calculate the molality (m).
To calculate molality, we use the formula:
molality (m) = moles of solute / mass of solvent (in kg)
Given:
mass of solvent = 95 g
Convert the mass of solvent to kilograms:
mass of solvent = 95 g / 1000 = 0.095 kg
Step 3: Calculate the boiling point elevation (∆Tb).
To calculate the boiling point elevation, we use the formula:
∆Tb = observed boiling point of the solution - boiling point of the pure solvent
Given:
observed boiling point of the solution = 56.50°C
boiling point of pure acetone = 55.95°C
∆Tb = 56.50°C - 55.95°C = 0.55°C
Step 4: Calculate the molar mass (M) of the solute.
Now, we can substitute the known values into the formula for boiling point elevation:
∆Tb = Kbm
Solving for the molar mass (M):
M = ∆Tb / (Kb * m)
Given:
Kb (molar boiling point elevation constant of acetone) = 1.71 °C kg/mol
Substitute the values:
M = 0.55°C / (1.71°C kg/mol * 0.095 kg)
M = 3.508 mol/kg
The molar mass of the solute is approximately 3.508 g/mol.