Answer :
Certainly! Let's go through the molarity calculations step by step for each solution.
### 1. Sodium Chloride (NaCl) Solution
- Given: 2.3 moles of NaCl in 0.45 liters of solution.
- Molarity formula: [tex]\( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \)[/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{2.3 \text{ moles}}{0.45 \text{ liters}} = 5.1111 \text{ M} \][/tex]
So, the molarity is approximately 5.1111 M.
### 2. Calcium Carbonate (CaCO[tex]\(_3\)[/tex]) Solution
- Given: 1.2 moles of CaCO[tex]\(_3\)[/tex] in 1.22 liters of solution.
- Calculation:
[tex]\[ \text{Molarity} = \frac{1.2 \text{ moles}}{1.22 \text{ liters}} = 0.9836 \text{ M} \][/tex]
So, the molarity is approximately 0.9836 M.
### 3. Sodium Sulfate (Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) Solution
- Given: 0.09 moles of Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] in 12 mL of solution.
- First, convert mL to liters:
[tex]\[ 12 \text{ mL} = 0.012 \text{ liters} \][/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.09 \text{ moles}}{0.012 \text{ liters}} = 7.5 \text{ M} \][/tex]
So, the molarity is 7.5 M.
### 4. Lithium Fluoride (LiF) Solution
- Given: 0.75 moles of LiF in 65 mL of solution.
- First, convert mL to liters:
[tex]\[ 65 \text{ mL} = 0.065 \text{ liters} \][/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.75 \text{ moles}}{0.065 \text{ liters}} = 11.5385 \text{ M} \][/tex]
So, the molarity is approximately 11.5385 M.
### 5. Magnesium Acetate (Mg(C[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex])[tex]\(_2\)[/tex]) Solution
- Given: 0.8 moles of Mg(C[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex])[tex]\(_2\)[/tex] in 5 liters of solution.
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.8 \text{ moles}}{5 \text{ liters}} = 0.16 \text{ M} \][/tex]
So, the molarity is 0.16 M.
### 6. Calcium Nitrite (Ca(NO[tex]\(_2\)[/tex])[tex]\(_2\)[/tex]) Solution
- Given: 120 grams of Ca(NO[tex]\(_2\)[/tex])[tex]\(_2\)[/tex] in 240 mL of solution.
- First, convert grams to moles using the molar mass of Ca(NO[tex]\(_2\)[/tex])[tex]\(_2\)[/tex], which is 132.14 grams per mole:
[tex]\[ \text{Moles of Ca(NO}_2\text{)}_2 = \frac{120 \text{ grams}}{132.14 \text{ grams/mole}} \approx 0.908 \text{ moles} \][/tex]
- Next, convert mL to liters:
[tex]\[ 240 \text{ mL} = 0.24 \text{ liters} \][/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.908 \text{ moles}}{0.24 \text{ liters}} = 3.7839 \text{ M} \][/tex]
So, the molarity is approximately 3.7839 M.
In summary, the molarities of each solution are:
1. Sodium Chloride: 5.1111 M
2. Calcium Carbonate: 0.9836 M
3. Sodium Sulfate: 7.5 M
4. Lithium Fluoride: 11.5385 M
5. Magnesium Acetate: 0.16 M
6. Calcium Nitrite: 3.7839 M
### 1. Sodium Chloride (NaCl) Solution
- Given: 2.3 moles of NaCl in 0.45 liters of solution.
- Molarity formula: [tex]\( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \)[/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{2.3 \text{ moles}}{0.45 \text{ liters}} = 5.1111 \text{ M} \][/tex]
So, the molarity is approximately 5.1111 M.
### 2. Calcium Carbonate (CaCO[tex]\(_3\)[/tex]) Solution
- Given: 1.2 moles of CaCO[tex]\(_3\)[/tex] in 1.22 liters of solution.
- Calculation:
[tex]\[ \text{Molarity} = \frac{1.2 \text{ moles}}{1.22 \text{ liters}} = 0.9836 \text{ M} \][/tex]
So, the molarity is approximately 0.9836 M.
### 3. Sodium Sulfate (Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]) Solution
- Given: 0.09 moles of Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] in 12 mL of solution.
- First, convert mL to liters:
[tex]\[ 12 \text{ mL} = 0.012 \text{ liters} \][/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.09 \text{ moles}}{0.012 \text{ liters}} = 7.5 \text{ M} \][/tex]
So, the molarity is 7.5 M.
### 4. Lithium Fluoride (LiF) Solution
- Given: 0.75 moles of LiF in 65 mL of solution.
- First, convert mL to liters:
[tex]\[ 65 \text{ mL} = 0.065 \text{ liters} \][/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.75 \text{ moles}}{0.065 \text{ liters}} = 11.5385 \text{ M} \][/tex]
So, the molarity is approximately 11.5385 M.
### 5. Magnesium Acetate (Mg(C[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex])[tex]\(_2\)[/tex]) Solution
- Given: 0.8 moles of Mg(C[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex])[tex]\(_2\)[/tex] in 5 liters of solution.
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.8 \text{ moles}}{5 \text{ liters}} = 0.16 \text{ M} \][/tex]
So, the molarity is 0.16 M.
### 6. Calcium Nitrite (Ca(NO[tex]\(_2\)[/tex])[tex]\(_2\)[/tex]) Solution
- Given: 120 grams of Ca(NO[tex]\(_2\)[/tex])[tex]\(_2\)[/tex] in 240 mL of solution.
- First, convert grams to moles using the molar mass of Ca(NO[tex]\(_2\)[/tex])[tex]\(_2\)[/tex], which is 132.14 grams per mole:
[tex]\[ \text{Moles of Ca(NO}_2\text{)}_2 = \frac{120 \text{ grams}}{132.14 \text{ grams/mole}} \approx 0.908 \text{ moles} \][/tex]
- Next, convert mL to liters:
[tex]\[ 240 \text{ mL} = 0.24 \text{ liters} \][/tex]
- Calculation:
[tex]\[ \text{Molarity} = \frac{0.908 \text{ moles}}{0.24 \text{ liters}} = 3.7839 \text{ M} \][/tex]
So, the molarity is approximately 3.7839 M.
In summary, the molarities of each solution are:
1. Sodium Chloride: 5.1111 M
2. Calcium Carbonate: 0.9836 M
3. Sodium Sulfate: 7.5 M
4. Lithium Fluoride: 11.5385 M
5. Magnesium Acetate: 0.16 M
6. Calcium Nitrite: 3.7839 M
