a solution containing 100g of unknown liquid and 900g water has a freezing point pf -3.33 C. given kf=1.86 degrees c for water, the moelcular weight of the unknown liquid is what?
delta T = Kf*molality
Solve for m
m = #moles/kg solvent
Substitute m and kg solvent, solve for # moles.
# moles = grams/molar mass
You have #moles and grams, solve for molar mass.
To find the molecular weight of the unknown liquid, we need to use the concept of freezing point depression. Freezing point depression occurs when a solute (in this case, the unknown liquid) is dissolved in a solvent (water), causing the freezing point of the solvent to decrease.
The formula for freezing point depression is:
ΔT = Kf * m
Where:
ΔT = Change in freezing point
Kf = Freezing point depression constant for the solvent (water in this case)
m = Molality of the solvent
First, let's calculate the molality (m) of the water solution.
Molality (m) = moles of solute / mass of solvent (in kg)
The mass of the water in kilograms is 900g / 1000 = 0.9 kg (since 1 kg = 1000g).
Next, we calculate the moles of the solute (unknown liquid).
Moles of solute = Mass of solute / Molecular weight of solute
Given that the mass of the solute is 100g and let's assume the molecular weight of the unknown liquid as "M," we can rewrite the equation as:
Moles of solute (unknown liquid) = 100g / M g/mol
Now, we substitute these values into the freezing point depression equation:
ΔT = Kf * m
-3.33°C = 1.86°C/m * (100g / M g/mol) * (0.9 kg)
Simplifying the equation:
-3.33°C = 1.674 / M mol/kg
To isolate M, we can rearrange the equation:
M = 1.674 / (-3.33°C * 0.9 mol/kg)
Calculating further:
M = -1.674 / (-3.33 * 0.9)
M ≈ 1.674 / 2.997
M ≈ 0.56 g/mol
Therefore, the molecular weight of the unknown liquid is approximately 0.56 g/mol.