posted by cateye on .
You dissolve 6.55 g of potassium nitrate in enough water to make a 250. ML solution.
1.What is the molarity of this solution?answer:0.0262
2.If you took out 15.0 mL of this solution, how many moles of potassium nitrate would you have?answer:0.393
3. What is the percent (m/v) of this solution?answer:0.0262
4.If the resulting solution has a density of 1.45 g/mL, what is the percent (m/m) of the solution?answer:0.0181
5.How does the freezing point of this solution compare with that of pure water?answer:should be lower
6.How does the boiling point compare?answer:it should be higher than that for pure water
7.If you added 1.25 L of additional water to this solution, what would the molarity of this new solution be?answer:.0432
4,5,6 look ok.
1,2,and 3 I can't confirm.
Show your work on those.
I didn't check 7.
This is what I did..please check it
1, 2, and 3 are NOT gas law problems. They are concentration problems.
mols = grams/molar mass
Then M = mols/L soln
mols = M x L = ?
3. You have 6.55 g in 250 mL. Convert this to g/100 for % m/v.
DID I DO IT RIGHT NOW
2.m=6.55g*15.0=.89 so this mean I need to use the full 250ml
correctly for #2 is 98.25
1. My calculator reads 0.25941 which I would round to 0.259 and not 0.260. The reason you obtained 0.260 is because you rounded incorrectly for the first part of the question; i.e., 6.55/101 = 0.06485. If you round that off too much that makes all of the remaining answers that follow that wrong. This is what you should do.
6.55/101 = some number BUT LEAVE THAT IN THE CALCULATOR. I don't even read the number and that saves some time. Then divide by 0.250L. That way you don't make rounding errors. ;-).
The question is to find the number of mols in a 15.0 mL portion.
mols = M x L = 0.0259M x 0.015L = ?
3. 2.62 is correct but some profs get picky and will count off if you don't write the unit, also. I would write 2.62% m/v
7. Now that you have #1, #7 follows:
0.0259M x [250 mL/(250 mL + 1250 mL)] = ? or approximately 0.004 M