[pixyatplay]

How would one go from 1.546 g/mL of potassium iodide (MM 166) to molarity then to molality and then to mass percent of solute?  This is boggling my brain, is there anyone who could offer any assistance at all?  It would be greatly appreciated.


Thus far I have gotten that from a density of 1.546 g/mL one would have 6.021 m (molality) but could only get to this trhough the given of 4.654 M, could someone perhaps be able to assist me in showing the conversion there, so I could perform it on my own in the future.  I got 6.021 for molality, but am not sure how to go from all these numbers to mass percent.  Any assistance would help.  Thanks.

I got to molality by finding the grams of water 772.564, then using the 4.654 (which was a given) divided 4.654 moles by .773 kg which I reached by subtracting 772.564 from 1546 grams or the whole of solution getting 773.436g and converting to kg.  my solution therefore was 6.021.

[mdlhvn]

First, some definition:

d=density of solution, g/ml
C%= percent concentration of solution, means the mass (g) of solute per 100g solution.
Cm=molarity of solution, mole/L.
M: molecule mass of solute.

If the Cm is provided, mean Cm moles of solute in 1L(1000ml) solution, the we calculate:

-Mass of solute=Cm*M, set m1
-Mass of solution= d*V=d*1000, set it m0

C%=(m1/m0)*100%= Cm*M/d*10.

Similarily, you can calculate Cm from C% from the equation above, pay attention to the unit. The important thing is that you understand the meaning of each terms.

[Borek]

Concentrations explained

Concentration cheat sheet

[pixyatplay]

Thank you both very much for responding, you saved me, and shed a little light on the topic so I can practice it before my final.  Again thanks.