complete the following practice problems. answers are at the bottom of the page.
1. determine the concentration of 0.650 moles of solute in 2.8 l.
2. determine the concentration of 1.58 moles of solute in 750 ml.
3. determine the volume of solution if 3.85 moles are in a solution with a concentration of 14.3 m.
4. determine the number of moles in a 555 ml solution with a concentration of 3.75 m.
5. determine the number of moles in a 5.25 l solution with a concentration of 12.5 m.
6. how many grams of potassium carbonate are needed to make 200.0 ml of a 2.50 m solution?
7. how many liters of 4.0 m solution can be made using 100.0 grams of lithium bromide?
8. what is the concentration of an aqueous solution with a volume of 450. ml that contains 200.0 grams of iron (iii) chloride?
9. how many grams of copper (ii) fluoride are needed to make 6.70 l of a 1.20 m

Question

Accepted Answer

Let's solve problem 1 first: "Determine the concentration of 0.650 moles of solute in 2.8 L."

# Explanation:
## Step1: Recall the formula for molarity (M)
Molarity ($M$) is defined as moles of solute ($n$) divided by volume of solution in liters ($V$), so the formula is $M = \frac{n}{V}$.
## Step2: Substitute the given values
We have $n = 0.650$ moles and $V = 2.8$ L. Plugging these into the formula: $M = \frac{0.650}{2.8}$.
## Step3: Calculate the result
$\frac{0.650}{2.8} \approx 0.232$ M (rounded to three significant figures).

# Answer:
The concentration is approximately $\boldsymbol{0.232\ 	ext{M}}$.

Now problem 2: "Determine the concentration of 1.58 moles of solute in 750 mL."

# Explanation:
## Step1: Convert volume to liters
Since molarity uses liters, convert 750 mL to liters. There are 1000 mL in 1 L, so $V = \frac{750}{1000} = 0.75$ L.
## Step2: Use the molarity formula
$M = \frac{n}{V}$, with $n = 1.58$ moles and $V = 0.75$ L.
## Step3: Calculate the molarity
$M = \frac{1.58}{0.75} \approx 2.11$ M (rounded to three significant figures).

# Answer:
The concentration is approximately $\boldsymbol{2.11\ 	ext{M}}$.

Problem 3: "Determine the volume of solution if 3.85 moles are in a solution with a concentration of 14.3 M."

# Explanation:
## Step1: Rearrange the molarity formula for volume
From $M = \frac{n}{V}$, we can solve for $V$: $V = \frac{n}{M}$.
## Step2: Substitute the values
We have $n = 3.85$ moles and $M = 14.3$ M. So $V = \frac{3.85}{14.3}$.
## Step3: Calculate the volume
$\frac{3.85}{14.3} \approx 0.269$ L (or 269 mL, rounded to three significant figures).

# Answer:
The volume is approximately $\boldsymbol{0.269\ 	ext{L}}$ (or 269 mL).

Problem 4: "Determine the number of moles in a 555 mL solution with a concentration of 3.75 M."

# Explanation:
## Step1: Convert volume to liters
555 mL = $\frac{555}{1000} = 0.555$ L.
## Step2: Rearrange molarity formula for moles
From $M = \frac{n}{V}$, solve for $n$: $n = M 	imes V$.
## Step3: Substitute values
$M = 3.75$ M and $V = 0.555$ L. So $n = 3.75 	imes 0.555$.
## Step4: Calculate moles
$3.75 	imes 0.555 = 2.08125 \approx 2.08$ moles (rounded to three significant figures).

# Answer:
The number of moles is approximately $\boldsymbol{2.08\ 	ext{moles}}$.

Problem 5: "Determine the number of moles in a 5.25 L solution with a concentration of 12.5 M."

# Explanation:
## Step1: Use the formula $n = M 	imes V$
We know $M = 12.5$ M and $V = 5.25$ L.
## Step2: Calculate moles
$n = 12.5 	imes 5.25 = 65.625 \approx 65.6$ moles (rounded to three significant figures).

# Answer:
The number of moles is $\boldsymbol{65.6\ 	ext{moles}}$ (or 65.625 moles).

Problem 6: "How many grams of potassium carbonate are needed to make 200.0 mL of a 2.50 M solution?"

# Explanation:
## Step1: Find moles of $K_2CO_3$
First, convert volume to liters: 200.0 mL = 0.2000 L. Then use $n = M 	imes V$. So $n = 2.50\ 	ext{M} 	imes 0.2000\ 	ext{L} = 0.500$ moles.
## Step2: Find molar mass of $K_2CO_3$
Molar mass of K: 39.10 g/mol, C: 12.01 g/mol, O: 16.00 g/mol. So molar mass of $K_2CO_3$ is $2(39.10) + 12.01 + 3(16.00) = 78.20 + 12.01 + 48.00 = 138.21$ g/mol.
## Step3: Calculate mass
Mass = moles × molar mass = $0.500\ 	ext{mol} 	imes 138.21\ 	ext{g/mol} = 69.105\ 	ext{g} \approx 69.1$ g (rounded to three significant figures).

# Answer:
The mass of potassium carbonate needed is approximately $\boldsymbol{69.1\ 	ext{g}}$.

Problem 7: "How many liters of 4.0 M solution can be made using 100.0 grams of lithium bromide?"

# Explanation:
## Step1: Find moles of LiBr
Molar mass of Li: 6.94 g/mol, Br: 79.90 g/mol. So molar mass of LiBr is $6.94 + 79.90 = 86.84$ g/mol. Moles $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{100.0\ 	ext{g}}{86.84\ 	ext{g/mol}} \approx 1.1515$ moles.
## Step2: Use molarity formula to find volume
From $M = \frac{n}{V}$, solve for $V$: $V = \frac{n}{M}$. $M = 4.0$ M, so $V = \frac{1.1515}{4.0} \approx 0.288$ L (rounded to three significant figures).

# Answer:
The volume of the solution is approximately $\boldsymbol{0.288\ 	ext{L}}$.

Problem 8: "What is the concentration of an aqueous solution with a volume of 450. mL that contains 200.0 grams of iron (III) chloride?"

# Explanation:
## Step1: Find moles of $FeCl_3$
Molar mass of Fe: 55.85 g/mol, Cl: 35.45 g/mol. So molar mass of $FeCl_3$ is $55.85 + 3(35.45) = 55.85 + 106.35 = 162.2$ g/mol. Moles $n = \frac{200.0\ 	ext{g}}{162.2\ 	ext{g/mol}} \approx 1.233$ moles.
## Step2: Convert volume to liters
450. mL = 0.450 L.
## Step3: Calculate molarity
$M = \frac{n}{V} = \frac{1.233\ 	ext{mol}}{0.450\ 	ext{L}} \approx 2.74$ M (rounded to three significant figures).

# Answer:
The concentration is approximately $\boldsymbol{2.74\ 	ext{M}}$.

Problem 9: "How many grams of copper (II) fluoride are needed to make 6.70 L of a 1.20 M solution?"

# Explanation:
## Step1: Find moles of $CuF_2$
Using $n = M 	imes V$, $M = 1.20$ M, $V = 6.70$ L. So $n = 1.20 	imes 6.70 = 8.04$ moles.
## Step2: Find molar mass of $CuF_2$
Molar mass of Cu: 63.55 g/mol, F: 19.00 g/mol. So molar mass of $CuF_2$ is $63.55 + 2(19.00) = 63.55 + 38.00 = 101.55$ g/mol.
## Step3: Calculate mass
Mass = moles × molar mass = $8.04\ 	ext{mol} 	imes 101.55\ 	ext{g/mol} \approx 816$ g (rounded to three significant figures).

# Answer:
The mass of copper (II) fluoride needed is approximately $\boldsymbol{816\ 	ext{g}}$.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for molarity (M)

Step2: Substitute the given values

Step3: Calculate the result

Step1: Convert volume to liters

Step2: Use the molarity formula

Step3: Calculate the molarity

Step1: Rearrange the molarity formula for volume

Step2: Substitute the values

Step3: Calculate the volume

Answer: