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?
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.
