Calculate the molality, molarity and mole fraction of a $\ce{CuSO4}$ in a 17% m/m aqueous solution (density of the solution is 1.367 g/mole)

I did .17 * 1.367 to find grams of $\ce{CuSO4}$

and .83 * 1.367 to find grams of $\ce{H2O}$

Then used to them calculate:

molarity = 0.00146 M

molality = 1.287 m

Mole fraction = 0.023

I'm wondering if I'm doing it right?


1 Answer 1


First let me correct you, m/m means mass/mass % I think you assumed it to be mole/mole.

Assume you have 100g of solution in which Cu is 17g and solvent is 83 gm. 17 gm copper means 0.268 moles of Cu.

Now, density = Total weight/Total moles = 100/(0.268 + x) = 1.367 where x is moles of solvent, from here you will get x = 72.8849

Thus, mole fraction of Cu is 0.268/(0.268+72.8849) = 0.00366

Molality = Moles of solute/Weight of Solvent(in Kg) = 0.268/0.083 = 3.229 m

Since, density in terms of g/l is not given I don't think it's possible to calculate molarity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.