Answer :
To calculate the vapor pressure of the solution, we can use Raoult's Law, which states that the vapor pressure of a solvent over a solution (P_solution) is equal to the vapor pressure of the pure solvent (P_pure) multiplied by the mole fraction of the solvent (X_solvent).
Formula for Raoult's Law:
[tex]\[ P_{solution} = X_{solvent} \times P_{pure} \][/tex]
First, let's find the mole fraction of water in the solution:
1. Calculate the moles of NaCl and water in the solution.
Given that the NaCl solution is 5.50% by mass, for a hypothetical 100 g of the solution, we have:
- 5.50 g of NaCl
- 94.50 g of H2O
Now let's calculate moles for each component using their molecular weights:
- NaCl has a molecular weight of approximately [tex]\( 58.44 \, \text{g/mol} \)[/tex]
- H2O has a molecular weight of approximately [tex]\( 18.02 \, \text{g/mol} \)[/tex]
[tex]\[ \text{Moles of NaCl} = \frac{5.50 \, \text{g}}{58.44 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of NaCl} \approx 0.0941 \, \text{mol} \][/tex]
[tex]\[ \text{Moles of H2O} = \frac{94.50 \, \text{g}}{18.02 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of H2O} \approx 5.245 \, \text{mol} \][/tex]
2. Calculate the mole fraction of water.
The mole fraction of water (X_water) is given by the number of moles of water divided by the total number of moles in the solution.
[tex]\[ X_{water} = \frac{\text{Moles of H2O}}{\text{Moles of NaCl} + \text{Moles of H2O}} \][/tex]
[tex]\[ X_{water} = \frac{5.245}{0.0941 + 5.245} \][/tex]
[tex]\[ X_{water} = \frac{5.245}{5.3391} \][/tex]
[tex]\[ X_{water} \approx 0.9823 \][/tex]
3. Apply Raoult's Law to find the vapor pressure of the solution.
Now we can use Raoult's Law to find the vapor pressure of the solution. The vapor pressure of pure water at 25°C is given as 23.8 torr.
[tex]\[ P_{solution} = X_{water} \times P_{pure} \][/tex]
[tex]\[ P_{solution} = 0.9823 \times 23.8 \, \text{torr} \][/tex]
[tex]\[ P_{solution} \approx 23.38 \, \text{torr} \][/tex]
Therefore, the vapor pressure at 25°C of the aqueous solution that is 5.50% NaCl by mass is approximately 23.38 torr.
Formula for Raoult's Law:
[tex]\[ P_{solution} = X_{solvent} \times P_{pure} \][/tex]
First, let's find the mole fraction of water in the solution:
1. Calculate the moles of NaCl and water in the solution.
Given that the NaCl solution is 5.50% by mass, for a hypothetical 100 g of the solution, we have:
- 5.50 g of NaCl
- 94.50 g of H2O
Now let's calculate moles for each component using their molecular weights:
- NaCl has a molecular weight of approximately [tex]\( 58.44 \, \text{g/mol} \)[/tex]
- H2O has a molecular weight of approximately [tex]\( 18.02 \, \text{g/mol} \)[/tex]
[tex]\[ \text{Moles of NaCl} = \frac{5.50 \, \text{g}}{58.44 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of NaCl} \approx 0.0941 \, \text{mol} \][/tex]
[tex]\[ \text{Moles of H2O} = \frac{94.50 \, \text{g}}{18.02 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of H2O} \approx 5.245 \, \text{mol} \][/tex]
2. Calculate the mole fraction of water.
The mole fraction of water (X_water) is given by the number of moles of water divided by the total number of moles in the solution.
[tex]\[ X_{water} = \frac{\text{Moles of H2O}}{\text{Moles of NaCl} + \text{Moles of H2O}} \][/tex]
[tex]\[ X_{water} = \frac{5.245}{0.0941 + 5.245} \][/tex]
[tex]\[ X_{water} = \frac{5.245}{5.3391} \][/tex]
[tex]\[ X_{water} \approx 0.9823 \][/tex]
3. Apply Raoult's Law to find the vapor pressure of the solution.
Now we can use Raoult's Law to find the vapor pressure of the solution. The vapor pressure of pure water at 25°C is given as 23.8 torr.
[tex]\[ P_{solution} = X_{water} \times P_{pure} \][/tex]
[tex]\[ P_{solution} = 0.9823 \times 23.8 \, \text{torr} \][/tex]
[tex]\[ P_{solution} \approx 23.38 \, \text{torr} \][/tex]
Therefore, the vapor pressure at 25°C of the aqueous solution that is 5.50% NaCl by mass is approximately 23.38 torr.
