Posted by **Paul** on Thursday, February 10, 2011 at 12:54am.

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. ??????

- Chemistry- Drbob-help#3 -
**DrBob222**, Thursday, February 10, 2011 at 1:30am
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.

- Chemistry- Drbob-help#3 -
**Paul**, Thursday, February 10, 2011 at 2:00am
Thanks I understand now what I did wrong you have been a big help.

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