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Sagot :
Sure! Let's break this problem down step-by-step to calculate the mass of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex] needed to create a 100 ml of 0.755 M solution.
### Step 1: Determine the Volume in Liters
First, convert the volume from milliliters to liters because molarity (M) is defined in terms of liters of solution:
[tex]\[ \text{Volume} = 100 \, \text{ml} = 100 / 1000 \, \text{liters} = 0.1 \, \text{liters} \][/tex]
### Step 2: Calculate the Moles of Solute Needed
Next, use the molarity of the solution to find the number of moles of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex]:
[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]
Therefore, the moles of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex] needed is:
[tex]\[ \text{Moles} = 0.755 \, \text{M} \times 0.1 \, \text{liters} = 0.0755 \, \text{moles} \][/tex]
### Step 3: Calculate the Molar Mass of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex]
Now, sum the atomic masses of the constituent elements:
- Copper (Cu): [tex]\( 63.55 \, \text{g/mol} \)[/tex]
- Sulfur (S): [tex]\( 32.07 \, \text{g/mol} \)[/tex]
- Oxygen (O) in [tex]\( \text{O}_4 \)[/tex]: [tex]\( 4 \times 16.00 \, \text{g/mol} = 64.00 \, \text{g/mol} \)[/tex]
- Water in [tex]\( 5 \text{H}_2 \text{O} \)[/tex]: [tex]\( 5 \times (2 \times 1.01 \, \text{g/mol} + 16.00 \, \text{g/mol}) = 5 \times 18.02 \, \text{g/mol} = 90.10 \, \text{g/mol} \)[/tex]
Adding these together gives:
[tex]\[ \text{Molar Mass} = 63.55 + 32.07 + 64.00 + 90.10 = 249.72 \, \text{g/mol} \][/tex]
### Step 4: Calculate the Mass of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex] Needed
Finally, calculate the mass required using the moles and molar mass:
[tex]\[ \text{Mass} = \text{moles} \times \text{molar mass} \][/tex]
[tex]\[ \text{Mass} = 0.0755 \, \text{moles} \times 249.72 \, \text{g/mol} = 18.85 \, \text{grams} \][/tex]
This is the mass of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex] required to make a 100 ml solution with a molarity of 0.755 M.
### Step 1: Determine the Volume in Liters
First, convert the volume from milliliters to liters because molarity (M) is defined in terms of liters of solution:
[tex]\[ \text{Volume} = 100 \, \text{ml} = 100 / 1000 \, \text{liters} = 0.1 \, \text{liters} \][/tex]
### Step 2: Calculate the Moles of Solute Needed
Next, use the molarity of the solution to find the number of moles of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex]:
[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]
Therefore, the moles of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex] needed is:
[tex]\[ \text{Moles} = 0.755 \, \text{M} \times 0.1 \, \text{liters} = 0.0755 \, \text{moles} \][/tex]
### Step 3: Calculate the Molar Mass of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex]
Now, sum the atomic masses of the constituent elements:
- Copper (Cu): [tex]\( 63.55 \, \text{g/mol} \)[/tex]
- Sulfur (S): [tex]\( 32.07 \, \text{g/mol} \)[/tex]
- Oxygen (O) in [tex]\( \text{O}_4 \)[/tex]: [tex]\( 4 \times 16.00 \, \text{g/mol} = 64.00 \, \text{g/mol} \)[/tex]
- Water in [tex]\( 5 \text{H}_2 \text{O} \)[/tex]: [tex]\( 5 \times (2 \times 1.01 \, \text{g/mol} + 16.00 \, \text{g/mol}) = 5 \times 18.02 \, \text{g/mol} = 90.10 \, \text{g/mol} \)[/tex]
Adding these together gives:
[tex]\[ \text{Molar Mass} = 63.55 + 32.07 + 64.00 + 90.10 = 249.72 \, \text{g/mol} \][/tex]
### Step 4: Calculate the Mass of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex] Needed
Finally, calculate the mass required using the moles and molar mass:
[tex]\[ \text{Mass} = \text{moles} \times \text{molar mass} \][/tex]
[tex]\[ \text{Mass} = 0.0755 \, \text{moles} \times 249.72 \, \text{g/mol} = 18.85 \, \text{grams} \][/tex]
This is the mass of [tex]\( \text{CuSO}_4 \cdot 5 \text{H}_2 \text{O} \)[/tex] required to make a 100 ml solution with a molarity of 0.755 M.
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