If you dissolve 29.25 g of sodium chloride in 250 g of H2O. What would be the freezing point?
Kfp- -1.86 degrees C/molal
I know the formula is K times molality.
I am taking -1.86(2.00) and getting -3.72 degrees Celsius, when the answer is -7.44 degrees. What am I doing wrong?
delta T = i*Kf*m
Your problem is you have half the answer which means you aren't using i = van't Hoff factor.
How many moles NaCl in 29.25g? That will be moles = grams/molar mass. Solve for moles.
What is molality?
That will be m = moles/kg solvent
Solve for m
What is i?
i = 2 for Nacl.
Substitute into the formula.
To calculate the freezing point depression, you need to determine the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent.
First, we need to find the number of moles of sodium chloride (NaCl). To do this, we divide the mass of NaCl (29.25 g) by its molar mass (58.44 g/mol):
Moles of NaCl = 29.25 g / 58.44 g/mol = 0.500 mol
Next, calculate the mass of the water (H2O) by subtracting the mass of NaCl from the total mass of the solution:
Mass of H2O = Total mass of solution - Mass of NaCl
Mass of H2O = 250 g - 29.25 g = 220.75 g
Now, convert the mass of H2O to kilograms:
Mass of H2O = 220.75 g / 1000 g/kg = 0.22075 kg
Finally, you can calculate the molality (m) using the equation:
molality (m) = moles of NaCl / mass of H2O in kg
molality (m) = 0.500 mol / 0.22075 kg = 2.26 mol/kg (rounded to two decimal places)
Now, to calculate the freezing point depression, you multiply the molality by the cryoscopic constant (Kfp = -1.86 °C/molal):
Freezing point depression (ΔTf) = molality (m) x Kfp
ΔTf = 2.26 mol/kg x -1.86 °C/molal = -4.20 °C (rounded to two decimal places)
Therefore, the freezing point of the solution would be -4.20 °C (rounded to two decimal places), not -7.44 °C. It seems there was an error in the book or the given answer.