Answer :
To determine the mass of \( O_2 \) in the flask, we will use the ideal gas law, which is given by the equation:
[tex]\[ PV = nRT \][/tex]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin.
Let's go step-by-step through the solution:
1. Convert the pressure from mm Hg to atm:
The pressure \( P \) is given as \( 322 \) mm Hg. To convert this to atmospheres (atm), we use the conversion factor \( 1 \) atm \( = 760 \) mm Hg.
[tex]\[ P_{\text{atm}} = \frac{322 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.4237 \, \text{atm} \][/tex]
2. Convert the temperature from Celsius to Kelvin:
The temperature \( T \) is given as \( 44^{\circ}C \). To convert this to Kelvin, we add \( 273.15 \):
[tex]\[ T_{\text{K}} = 44 + 273.15 = 317.15 \, \text{K} \][/tex]
3. Substitute the known values into the ideal gas law to solve for \( n \) (moles of \( O_2 \)):
The volume \( V \) is \( 15.0 \, \text{L} \), the gas constant \( R \) is \( 0.08206 \, \text{L atm} / (\text{mol K}) \).
[tex]\[ PV = nRT \implies n = \frac{PV}{RT} \][/tex]
Substituting the values:
[tex]\[ n = \frac{(0.4237 \, \text{atm})(15.0 \, \text{L})}{(0.08206 \, \text{L atm} / \text{mol K})(317.15 \, \text{K})} \][/tex]
[tex]\[ n = 0.2442 \, \text{moles} \][/tex]
4. Find the mass of \( O_2 \):
The molar mass of \( O_2 \) (oxygen gas) is \( 2 \times 16 \, \text{g/mol} = 32 \, \text{g/mol} \).
[tex]\[ \text{Mass of} \, O_2 = n \times \text{molar mass of} \, O_2 \][/tex]
Substituting \( n \):
[tex]\[ \text{Mass of} \, O_2 = 0.2442 \, \text{moles} \times 32 \, \text{g/mol} = 7.82 \, \text{g} \][/tex]
Therefore, the mass of \( O_2 \) in the flask is \( \mathbf{7.82 \, \text{g}} \).
Thus, the correct answer is:
c. [tex]\( 7.82 \, \text{g} \)[/tex]
[tex]\[ PV = nRT \][/tex]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin.
Let's go step-by-step through the solution:
1. Convert the pressure from mm Hg to atm:
The pressure \( P \) is given as \( 322 \) mm Hg. To convert this to atmospheres (atm), we use the conversion factor \( 1 \) atm \( = 760 \) mm Hg.
[tex]\[ P_{\text{atm}} = \frac{322 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.4237 \, \text{atm} \][/tex]
2. Convert the temperature from Celsius to Kelvin:
The temperature \( T \) is given as \( 44^{\circ}C \). To convert this to Kelvin, we add \( 273.15 \):
[tex]\[ T_{\text{K}} = 44 + 273.15 = 317.15 \, \text{K} \][/tex]
3. Substitute the known values into the ideal gas law to solve for \( n \) (moles of \( O_2 \)):
The volume \( V \) is \( 15.0 \, \text{L} \), the gas constant \( R \) is \( 0.08206 \, \text{L atm} / (\text{mol K}) \).
[tex]\[ PV = nRT \implies n = \frac{PV}{RT} \][/tex]
Substituting the values:
[tex]\[ n = \frac{(0.4237 \, \text{atm})(15.0 \, \text{L})}{(0.08206 \, \text{L atm} / \text{mol K})(317.15 \, \text{K})} \][/tex]
[tex]\[ n = 0.2442 \, \text{moles} \][/tex]
4. Find the mass of \( O_2 \):
The molar mass of \( O_2 \) (oxygen gas) is \( 2 \times 16 \, \text{g/mol} = 32 \, \text{g/mol} \).
[tex]\[ \text{Mass of} \, O_2 = n \times \text{molar mass of} \, O_2 \][/tex]
Substituting \( n \):
[tex]\[ \text{Mass of} \, O_2 = 0.2442 \, \text{moles} \times 32 \, \text{g/mol} = 7.82 \, \text{g} \][/tex]
Therefore, the mass of \( O_2 \) in the flask is \( \mathbf{7.82 \, \text{g}} \).
Thus, the correct answer is:
c. [tex]\( 7.82 \, \text{g} \)[/tex]
