Chemistry Drbobhelp
posted by Paul on .
I have placed the answer I got please help.
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. ??????

I have placed the answer I got please help.
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
ok
2. Calculate the % concentration of this solution. 55.8/930.8 *100=5.99%
OK
3. Calculate the molar concentration of this solution. ??930.8g/74.02=12.57/.870=14.45M I orginally got 1.14M
I think 1.14 is correct.
4. How many grams of potassium hydroxide are there in 150.0ml of this solution? 9.62g
moles = M x L.
moles KOH = 1.14 x 0.150 = ??
g KOH = moles x molar mass = ??
I get 9.59. Check my work. I used 56.1 for molar mass KOH.
1.14*0.15*56.1 = 9.59 g.
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
You method is ok but 1.14 for M
6. What volume of this solution contains 12.75 grams of potassium hydroxide? ??? .0393L
moles KOH in 12.75 g is 12.75/56.1 = ??
M = moles/L and rearrange to L = moles/M. Then L = ??moles/1.14 = xx
I'm a little confused by the term "this' solution. I don't know if it's referring to the original solution or the diluted solution. I've worked it for the original.
7. What mass of the original solution contains 25.0 grams of potassium hydroxide? 55.8g/930.8g=25/x=417g
OK
8. What volume of 0.750M H2SO4 solution can be neutralized by 100.0 grams of this solution? 1.2L
100 g of the solution contains 5.99 g of KOH. That is 5.99/56.1 = ??moles and it will neutralize 1/2 that moles of H2SO4. Then M = moles/L and rearrange to L = moles/M = ??moles/0.750 = zz L.
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. ??????
I would use mL x M = mL x M
mL x 1.14M = 600 mL x 0.1M and solve for mL of the original solution. I don't know the purpose of this problem; i.e., I may have missed something with the 1000 mL graduated cylinder. I think the volume needed is a little over 50 mL and that would be tough to measure very accurately in a 1000 mL graduated cylinder.