What is the molar mass of the unknown?

[huybinhs]

Homework Statement

A 1.40 g sample of an unknown is dissolved in 8.10 g of lauric acid. The freezing point is depressed 1.23 oC. What is the molar mass of the unknown?

Homework Equations

Delta T = Kf * m

The Attempt at a Solution

Delta T = Kf * m
=> m = 1.23 / 3.9 = 0.32

mole solute = 0.32 * 0.0081 = 0.0026 moles.

=> Molar mass = 1.40 / 0.0026 = 540 g/mol = final answer.

Am I doing correct? Please let me know!

 

[Borek]

Not bad. Two thngs that will make your solution better.

1. Don't round down intermediate results (check that it changes molar mass by about 8g).
2. Don't ignore units in your calculations, they are much easier to follow when it is obvious what is what of what.

 

[huybinhs]

Not bad. Two thngs that will make your solution better.

1. Don't round down intermediate results (check that it changes molar mass by about 8g).
2. Don't ignore units in your calculations, they are much easier to follow when it is obvious what is what of what.

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods

Thanks so much for your response!

I have edited my answer to:

Delta T = Kf * m => m = 1.23 / 3.9 = 0.32 m

mole solute = 0.32 * 0.0081 = 0.0026 mole

Molar mass of the unknown = 1.40 / 0.0026 = 538.46 g/mol .

Hope it better ;)

 

[Borek]

This time you overdid - you don't have enough information to give 5 significant digits in the answer But you see that rounding errors were responsible for 538 being 530.

 

[DeeNos]

Your approach is correct. To find the molar mass of the unknown, you first need to calculate the molality (m) of the solution using the given freezing point depression (ΔT) and the freezing point depression constant (Kf) for lauric acid. Then, you can use the equation m = moles of solute / mass of solvent to find the moles of solute (which is the unknown substance). Finally, divide the mass of the unknown substance by the number of moles to get the molar mass. Your final answer of 540 g/mol is correct. Good job!