Question
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. ??????
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. ??????
Answers
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
<b> ok</b>
2. Calculate the % concentration of this solution. 55.8/930.8 *100=5.99%
<b>OK</b>
3. Calculate the molar concentration of this solution. ??930.8g/74.02=12.57/.870=14.45M I orginally got 1.14M
<b>I think 1.14 is correct.</b>
4. How many grams of potassium hydroxide are there in 150.0ml of this solution? 9.62g
<b>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.</b>
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
<b>You method is ok but 1.14 for M</b>
6. What volume of this solution contains 12.75 grams of potassium hydroxide? ??? .0393L
<b>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.</b>
7. What mass of the original solution contains 25.0 grams of potassium hydroxide? 55.8g/930.8g=25/x=417g
<b>OK</b>
8. What volume of 0.750M H2SO4 solution can be neutralized by 100.0 grams of this solution? 1.2L
<b>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.</b>
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. ??????
<b>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. </b>
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
<b> ok</b>
2. Calculate the % concentration of this solution. 55.8/930.8 *100=5.99%
<b>OK</b>
3. Calculate the molar concentration of this solution. ??930.8g/74.02=12.57/.870=14.45M I orginally got 1.14M
<b>I think 1.14 is correct.</b>
4. How many grams of potassium hydroxide are there in 150.0ml of this solution? 9.62g
<b>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.</b>
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
<b>You method is ok but 1.14 for M</b>
6. What volume of this solution contains 12.75 grams of potassium hydroxide? ??? .0393L
<b>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.</b>
7. What mass of the original solution contains 25.0 grams of potassium hydroxide? 55.8g/930.8g=25/x=417g
<b>OK</b>
8. What volume of 0.750M H2SO4 solution can be neutralized by 100.0 grams of this solution? 1.2L
<b>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.</b>
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. ??????
<b>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. </b>