A compound containing only boron, nitrogen, and hydrogen was found to be 40.3% B, 5202% N, 7.5% H by mass. If 3.301g of this compound is dissolved in 50.00g of benzene, the solution produced freezes at 1.30 degree celsius. if the freezing point depression constant/cryoscopic constant is 5.12 degree celsius per metre and the freezing point of pure benzene is 5.48 degree celsius, what is the molecular weight of this compound?
please help! thanks in advance
This is a colligative property problem. Note that when an solute is dissolved in a solvent, the freezing point of the solvent decreases. This is called freezing depression, and the change in temperature can be calculated by
(T,orig - T,new) = Kf * m
where
T,orig = original freezing point of solvent
T,new = new freezing point
Kf = freezing point constant
m = molality (mol solute / kg solvent)
In the problem,
Freezing Point of Benzene = 5.48 C
The Kf for benzene = 5.12 C / m (I think in the problem, m is not 'meter' but it's molality)
We're looking for MW of the compound. Substituting,
(T,orig - T,new) = Kf * m
5.48 - 1.3 = 5.12 * (3.301 g / MW) / 0.05 kg benzene
4.18 * 0.05 = 16.901 / MW
MW = 16.901 / (4.18 * 0.05)
MW = 80.87 g / mol
Hope this helps :3
To determine the molecular weight of the given compound, we can use the freezing point depression formula:
ΔT = Kf * m
Where:
ΔT is the change in freezing point (in this case, ΔT = 5.48°C - 1.30°C = 4.18°C),
Kf is the freezing point depression constant (5.12°C/m),
and m is the molality of the solution.
First, calculate the molality (m) of the solution:
Moles of solute = mass of compound / molar mass of compound
Moles of solvent = mass of benzene / molar mass of benzene
Using the given percentages, assume we have a 100g sample of the compound:
- Boron (B) = 40.3g
- Nitrogen (N) = 52.02g
- Hydrogen (H) = 7.5g
Molar mass of B = 10.81 g/mol
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of compound = Molar mass of B + Molar mass of N + Molar mass of H
Calculate the moles of solute:
Moles of solute = (40.3g / 10.81 g/mol) + (52.02g / 14.01 g/mol) + (7.5g / 1.01 g/mol)
Next, calculate the moles of solvent:
Moles of solvent = mass of benzene / molar mass of benzene
Molar mass of benzene = 78.11 g/mol
Moles of solvent = 50.00g / 78.11 g/mol
Now, calculate the molality:
Molality (m) = Moles of solute / kg of solvent
Convert the mass of the solvent from grams to kilograms:
Mass of solvent = 50.00g / 1000 = 0.0500 kg
Molality (m) = Moles of solute / (Mass of solvent) in kg
Finally, plug the values into the freezing point depression equation and solve for the molecular weight (M):
ΔT = Kf * m
4.18°C = 5.12°C/m * m
From here, we can simplify the equation:
4.18°C = 5.12°C * m
m = 4.18°C / 5.12°C
Now, we can calculate the molecular weight (M):
Molecular weight (M) = moles of solute / moles of solvent
I hope this explanation helps you solve the problem!