A 33.4 g sample of a nonelectrolyte was dissolved is 680. g of water. The solution's freezing point was -2.89°C. What is the molar mass of the compound?
delta T = Kf*molality
2.89 = 1.86*m
Solve for m.
Then m = mols/kg solvent. You know m and kg solvent, substitute and solve for mols.
Then mols = g/molar mass. You know mols and g, solve for molar mass.
Post your work if you get stuck.
To find the molar mass of the compound, you need to use the concept of freezing point depression. The freezing point depression is directly related to the molality of the solution and the molal freezing point depression constant, which is a characteristic property of the solvent.
The equation for freezing point depression is given by:
ΔTf = Kf * m
where ΔTf is the change in freezing point, Kf is the molal freezing point depression constant of the solvent (for water, it is 1.86°C·kg/mol), and m is the molality of the solution.
In this case, we know the ΔTf (-2.89°C) and the mass of the solvent (water) in kilograms (680 g = 0.680 kg). We need to calculate the molality (m) of the solution.
The molality (m) is calculated using the formula:
m = (moles of solute) / (mass of solvent in kg)
First, we need to determine the moles of solute.
Given that the sample mass is 33.4 g, we can convert it to moles using the formula:
moles = mass (in grams) / molar mass
Now we can calculate the moles of solute.
moles = 33.4 g / molar mass
Substituting this into the equation for molality, we get:
m = (33.4 g / molar mass) / 0.680 kg
Simplifying further:
m = 33.4 g / (0.680 kg * molar mass)
Now we can substitute the values into the freezing point depression equation:
-2.89°C = (1.86°C·kg/mol) * (33.4 g / (0.680 kg * molar mass))
Now, let's solve for the molar mass.
Rearranging the equation, we have:
molar mass = 33.4 g / (1.86°C·kg/mol * -2.89°C) * 0.680 kg
molar mass = 33.4 g / (-2.89 * 1.86 * 0.680) kg·mol/°C
Calculating this gives us the molar mass of the compound.