posted by Paul on .
I still need help with #3 I'm not sure how I got 1.14M. I have tried to do the problem over, but I not coming up with my original answer of 1.14M. Somehow I am missing a step. Please show work for the 1.14M Thanks Question#9 reads exactly as written.
A student makes a solution by dissolving 55.8 grams of potassium hydroxide in 875.0 grams of water. The resulting solution has a density of 1.07 grams per milliliter.
1. What is the volume of this solution? 870mL
2. Calculate the % concentration of this solution. 55.8/930.8 *100=5.99%
3. Calculate the molar concentration of this solution. ??930.8g/74.02=12.57/.870=14.45M I orginally got 1.14M
4. How many grams of potassium hydroxide are there in 150.0ml of this solution? 9.62g
5. A 200.0ml sample of this solution is diluted to 500.0ml with water. What is its molar concentration?
14.45/x= 500ml/200ml= 5.78M
6. What volume of this solution contains 12.75 grams of potassium hydroxide? ??? .0393L
7. What mass of the original solution contains 25.0 grams of potassium hydroxide? 55.8g/930.8g=25/x=417g
8. What volume of 0.750M H2SO4 solution can be neutralized by 100.0 grams of this solution? 1.2L
9. A student needs 0.600L a 0.100M solution of potassium hydroxide. Describe how this can be done using the original solution and a 1000ml graduated cylinder. ??????
moles KOH = 55.8/56.1 = 0.99465
M = moles/L soln = 0.99465/0.870 L = 1.14328 M which rounds to 1.14M
If I'm missing something on #9 I don't see it. But I reiterate that 50 something mL would be difficult to measure accurately with a 1000 mL graduated cylinder.
Thanks I understand now what I did wrong you have been a big help.