| 1. |
A 0.750 L aqueous solution contains 90.0 g of ethanol, C2H5OH. Calculate the molar concentration of the solution in mol·L-1. |
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Solution:
| 1. |
The question asks for concentration,
which means finding molarity, or: |
moles solute
litre solution |
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| 2. |
To convert mass of ethanol to moles, we need to find the molar mass of C2H5OH using the periodic table. Molar mass is 46.1 g·mol-1 |
| 3. |
Molarity also requires volume; the question tells us we have 0.750 L. |
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Put this information together to solve the problem, arranging the information to end up with the desired unit:
mol
L |
= |
90.0 g |
× |
1 mol
46.1 g |
× |
1
0.750 L |
= |
2.60 mol
L |
Our final answer: [C2H5OH ] = 2.60M
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| 2. |
What mass of NaCl are dissolved in 152 mL of a solution if the concentration of the solution is 0.364 M? |
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Solution:
| 1. |
The question asks for mass, so we want to calculate grams
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| 2. |
We are given the concentration. I suggest you rewrite
the concentration as shown to the right, to better see
how the units will cancel out. |
0.364 mol
L |
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| 3. |
Because the question involves mass, we will need to know the molar mass of NaCl
Using a periodic table we find the molar mass of NaCl to be 58.5 g·mol-1 |
| 4. |
The question gives us the volume in mL. Our unit of concentration uses L, so we will convert 152 mL into 0.152 L. |
Put this information together to solve the problem, arranging the information to end up with the desired unit:
g |
= |
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× |
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× |
0.152 L |
= |
3.24 g |
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answer |
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| 3. |
What mass of dextrose, C6H12O6 is dissolved in 325 mL of 0.258 M solution? |
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Solution:
| 1. |
The question asks for mass, so we want to calculate grams
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| 2. |
We are given the concentration (0.258 M). I suggest you rewrite
the concentration as shown to the right, to better see
how the units will cancel out. |
0.258 mol
L |
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| 3. |
Because the question involves mass, we will need to know the molar mass of C6H12O6
Using a periodic table we find the molar mass of C6H12O6 to be 180.1 g·mol-1 |
| 4. |
The question gives us the volume in mL. Our unit of concentration uses L, so we will convert 325 mL into 0.325 L. |
Put this information together to solve the problem, arranging the information to end up with the desired unit:
g |
= |
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× |
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× |
0.325 L |
= |
15.1 g |
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answer |
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| 4. |
A mass of 98 g of sulfuric acid, H2SO4, is dissolved in water to prepare a 0.500 M solution. What is the volume of the solution? |
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Solution:
| 1. |
The question asks for volume, so we want to calculate litres, L (or mL)
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| 2. |
The concentration of the solution is: |
0.500 mol
L |
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| 3. |
Because the question involves mass, we will need to know the molar mass of H2SO4
Using a periodic table we find the molar mass of H2SO4 to be 98.1g·mol-1 |
Put this information together to solve the problem, arranging the information to end up with the desired unit:
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| 5. |
A solution of sodium carbonate, Na2CO3, contains 53.0 g of solute in 215 mL of solution. What is its molarity? |
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Solution:
| 1. |
| The question asks for molarity: |
moles solute
litre solution |
|
| 2. |
To convert mass of ethanol to moles, we need to find the molar mass of Na2CO3 using the periodic table. The molar mass of Na2CO3 is 106.0 g·mol-1 |
| 3. |
Molarity also requires volume; the question tells us we have 215 mL, or 0.215 L. |
Put this information together to solve the problem, arranging the information to end up with the desired unit:
mol
L |
= |
53.0 g |
× |
1 mol
106.0 g |
× |
1
0.215 L |
= |
2.33 mol
L |
Our final answer: [Na2CO3 ] = 2.33 M
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| 6. |
What is the molarity of a solution of HNO3 that contains 12.6 g of solute in 5.00 L of solution? |
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Solution:
| 1. |
| The question asks for molarity: |
moles solute
litre solution |
|
| 2. |
To convert mass of ethanol to moles, we need to find the molar mass of HNO3 using the periodic table. Molar mass is 64.0 g·mol-1 |
| 3. |
Molarity also requires volume; the question tells us we have 5.00 L. |
Put this information together to solve the problem, arranging the information to end up with the desired unit:
mol
L |
= |
12.6 g |
× |
1 mol
64.0 g |
× |
1
5.00 L |
= |
0.0393 mol
L |
Final answer: [HNO3] = 3.93 × 10-2 M
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| 7. |
What mass of copper(II) nitrate, Cu(NO3)2, is present in 50.00 mL of a 4.55 × 10-3 M aqueous solution? |
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Solution:
| 1. |
The question asks for mass, so we need to calculate grams
|
| 2. |
We are given the concentration. I suggest you rewrite
the concentration as shown to the right, to better see
how the units will cancel out. |
4.55 × 10-3 mol
L |
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| 3. |
Because the question involves mass, we will need to know the molar mass of Cu(NO3)2
Using a periodic table we find the molar mass of Cu(NO3)2 to be 187.6 g·mol-1 |
| 4. |
The question gives us the volume in mL. Our unit of concentration uses L, so we will convert 50.00 mL into 0.0.05000 L. |
Put this information together to solve the problem, arranging the information to end up with the desired unit:
g |
= |
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× |
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× |
0.0500 L |
= |
4.27 × 10-2 g |
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answer |
Final answer: 4.27 × 10-2 g of copper(II) nitrate are present.
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