Answer :
To determine how many grams of Sr(NO₃)₂ would be used in the preparation of 2.50 liters of a 3.5 M solution, we should proceed through the following steps:
1. Calculate the Molar Mass of Sr(NO₃)₂:
The molecular formula for Strontium nitrate is Sr(NO₃)₂.
- The atomic weight of Strontium (Sr) is 87.62 g/mol.
- The atomic weight of Nitrogen (N) is 14.01 g/mol.
- The atomic weight of Oxygen (O) is 16.00 g/mol.
Each nitrate ion contains one nitrogen atom and three oxygen atoms. Since there are two nitrate ions in Sr(NO₃)₂, the contributions from nitrogen and oxygen atoms are calculated as follows:
- Molecular weight of two nitrate ions (2 x (14.01 + 3 x 16.00)):
[tex]\[ 2 \times (14.01 + 3 \times 16.00) = 2 \times (14.01 + 48.00) = 2 \times 62.01 = 124.02 \, \text{g/mol} \][/tex]
- Adding the atomic weight of Sr:
[tex]\[ 87.62 \, \text{g/mol} + 124.02 \, \text{g/mol} = 211.64 \, \text{g/mol} \][/tex]
Thus, the molar mass of Sr(NO₃)₂ is 211.64 g/mol.
2. Calculate the Number of Moles of Sr(NO₃)₂ in the Solution:
The molarity (M) of the solution is given as 3.5 M, which represents 3.5 moles of solute per liter of solution. The volume of the solution is 2.50 liters.
- Number of moles ([tex]\(n\)[/tex]) of Sr(NO₃)₂ is calculated using the formula:
[tex]\[ n = \text{molarity} \times \text{volume (in liters)} \][/tex]
[tex]\[ n = 3.5 \, \text{M} \times 2.50 \, \text{L} = 8.75 \, \text{moles} \][/tex]
3. Calculate the Mass of Sr(NO₃)₂ Used:
The mass (m) of Sr(NO₃)₂ can be found by multiplying the number of moles by the molar mass:
[tex]\[ m = \text{moles} \times \text{molar mass} \][/tex]
[tex]\[ m = 8.75 \, \text{moles} \times 211.64 \, \text{g/mol} = 1851.85 \, \text{g} \][/tex]
Thus, 1851.85 grams of Sr(NO₃)₂ would be used in the preparation of 2.50 liters of a 3.5 M solution.
1. Calculate the Molar Mass of Sr(NO₃)₂:
The molecular formula for Strontium nitrate is Sr(NO₃)₂.
- The atomic weight of Strontium (Sr) is 87.62 g/mol.
- The atomic weight of Nitrogen (N) is 14.01 g/mol.
- The atomic weight of Oxygen (O) is 16.00 g/mol.
Each nitrate ion contains one nitrogen atom and three oxygen atoms. Since there are two nitrate ions in Sr(NO₃)₂, the contributions from nitrogen and oxygen atoms are calculated as follows:
- Molecular weight of two nitrate ions (2 x (14.01 + 3 x 16.00)):
[tex]\[ 2 \times (14.01 + 3 \times 16.00) = 2 \times (14.01 + 48.00) = 2 \times 62.01 = 124.02 \, \text{g/mol} \][/tex]
- Adding the atomic weight of Sr:
[tex]\[ 87.62 \, \text{g/mol} + 124.02 \, \text{g/mol} = 211.64 \, \text{g/mol} \][/tex]
Thus, the molar mass of Sr(NO₃)₂ is 211.64 g/mol.
2. Calculate the Number of Moles of Sr(NO₃)₂ in the Solution:
The molarity (M) of the solution is given as 3.5 M, which represents 3.5 moles of solute per liter of solution. The volume of the solution is 2.50 liters.
- Number of moles ([tex]\(n\)[/tex]) of Sr(NO₃)₂ is calculated using the formula:
[tex]\[ n = \text{molarity} \times \text{volume (in liters)} \][/tex]
[tex]\[ n = 3.5 \, \text{M} \times 2.50 \, \text{L} = 8.75 \, \text{moles} \][/tex]
3. Calculate the Mass of Sr(NO₃)₂ Used:
The mass (m) of Sr(NO₃)₂ can be found by multiplying the number of moles by the molar mass:
[tex]\[ m = \text{moles} \times \text{molar mass} \][/tex]
[tex]\[ m = 8.75 \, \text{moles} \times 211.64 \, \text{g/mol} = 1851.85 \, \text{g} \][/tex]
Thus, 1851.85 grams of Sr(NO₃)₂ would be used in the preparation of 2.50 liters of a 3.5 M solution.
