A solution was made by dissolving3.75g of a pure non-volatile solute in 95g of
acetone. The boiling point of pure acetone was observed to be 55.95°C, and that
of the solution was 56.S0°C. Ifthe molar boiling point elevation constant (2.43) of
acetone is 1.71°C kg/mol, what is the approximate molar mass of the solute?
To find the approximate molar mass of the solute, we can use the formula for boiling point elevation:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point elevation constant, and
m is the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.
To start, we need to calculate the molality of the solution. We can do this by first converting the mass of solute to moles, and then dividing by the mass of the solvent in kilograms.
First, let's convert the mass of the solute from grams to moles:
moles of solute = mass of solute / molar mass of solute
Since the mass of the solute is given as 3.75g, let's assume the molar mass of the solute is M.
moles of solute = 3.75g / M
Next, let's convert the mass of the solvent from grams to kilograms:
mass of solvent = 95g = 0.095kg
Now, we can calculate the molality of the solution:
molality (m) = moles of solute / mass of solvent in kg
m = (3.75g / M) / 0.095kg
Now, we can use the formula for boiling point elevation to find the change in boiling point (ΔTb):
ΔTb = Kb * m
Given that ΔTb is found to be approximately 0.55°C (56.50°C - 55.95°C), and the molal boiling point elevation constant (Kb) is 1.71°C kg/mol, we can rearrange the formula to solve for molar mass (M):
M = (3.75g / (0.095kg * 1.71°C/mol)) * 0.55°C
Simplifying the equation gives us:
M = 3.75g / (0.095kg * 1.71) * 0.55
Calculating the expression on the right side will give us the approximate molar mass of the solute.