For a.
Assume you have 1 mole of solution; therefore, you will have 0.0195 moles FeCl3 and 0.9805 moles H2O.
0.9805 moles x molar mass = 17.663 g H2O and since molality = moles/kg solvent that will be 0.0195 moles/0.017663 = ??m. Round to 3 s.f.
Check my arithmetic.
Calculate the concentrations of each of the following solutions.
(a) the molality of chloride ions in an aqueous solution of iron(III) chloride for which xFeCl3 is 0.0195
(b) the molality of hydroxide ions in a solution prepared from 9.16 g of barium hydroxide dissolved in 179 g of water
(c) the molality of 12.70 M NH3(aq) with a density of 0.9519 g·cm-3
I have repeatedly tried to figure out how to solve these, but I am getting something wrong. I know molality is mol solute/kg solvent. I'm especially having trouble with the second question. Please don't just give me the answer--try to explain how you solved it. An answer would be helpful so I can compare what I come up with though. Thanks!!
4 answers
Got it. for A i was using molar mass for FeCl3 instead of water..oops. I solved C...but I am still confused with the second question of anyone can help me out... Thanks!
the answer for a was actually incorrect
The answer to a is not incorrect if you re-read the problem. What you have calculated is molality of FeCl3, the problem asks for molality of Cl^-, so multiply the final answer you had by 3.
For #2, I think you reposted this later and I responded. If you still have questions or think there was an error, please repost a new post at the top of the page and I'll take a look at it.
For #2, I think you reposted this later and I responded. If you still have questions or think there was an error, please repost a new post at the top of the page and I'll take a look at it.
1 answer