Answer :
To determine how many grams of diphenyl (C₁₂H₁₀) must be dissolved in 50.0 grams of benzene to lower the freezing point by 2.00 °C, we can follow these steps:
### Step 1: Identify the given information
- Mass of benzene ([tex]\(m_{\text{benzene}}\)[/tex]) = 50.0 grams
- Freezing point depression ([tex]\(\Delta T\)[/tex]) = 2.00 °C
- Freezing point depression constant for benzene ([tex]\(K_f\)[/tex]) = 5.12 °C/m
- Molar mass of diphenyl (C₁₂H₁₀) = 154.21 g/mol
### Step 2: Calculate the molality of the solution
The freezing point depression ([tex]\(\Delta T\)[/tex]) is related to the molality (m) of the solution by the equation:
[tex]\[ \Delta T = K_f \cdot m \][/tex]
Rearranging the equation for molality (m):
[tex]\[ m = \frac{\Delta T}{K_f} \][/tex]
Substituting the known values:
[tex]\[ m = \frac{2.00 \, °C}{5.12 \, °C/m} \][/tex]
[tex]\[ m = 0.390625 \, m \][/tex]
This means the molality of the solution is 0.390625 mol/kg.
### Step 3: Calculate the moles of diphenyl required
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We can use this relationship to find the number of moles of diphenyl that need to be dissolved:
[tex]\[ m = \frac{\text{moles of diphenyl}}{ \text{mass of benzene in kg}} \][/tex]
We need to convert the mass of benzene into kilograms:
[tex]\[ \text{mass of benzene} = 50.0 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.0500 \, \text{kg} \][/tex]
Now, solving for the moles of diphenyl:
[tex]\[ \text{moles of diphenyl} = m \times \text{mass of benzene in kg} \][/tex]
[tex]\[ \text{moles of diphenyl} = 0.390625 \, m \times 0.0500 \, \text{kg} \][/tex]
[tex]\[ \text{moles of diphenyl} = 0.01953125 \, \text{mol} \][/tex]
### Step 4: Calculate the mass of diphenyl needed
Finally, we use the molar mass of diphenyl to convert the moles of diphenyl to grams:
[tex]\[ \text{mass of diphenyl} = \text{moles of diphenyl} \times \text{molar mass of diphenyl} \][/tex]
[tex]\[ \text{mass of diphenyl} = 0.01953125 \, \text{mol} \times 154.21 \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of diphenyl} = 3.0119140625 \, \text{g} \][/tex]
### Conclusion
To lower the freezing point of 50.0 grams of benzene by 2.00 °C, you must dissolve approximately 3.01 grams of diphenyl (C₁₂H₁₀) in it.
### Step 1: Identify the given information
- Mass of benzene ([tex]\(m_{\text{benzene}}\)[/tex]) = 50.0 grams
- Freezing point depression ([tex]\(\Delta T\)[/tex]) = 2.00 °C
- Freezing point depression constant for benzene ([tex]\(K_f\)[/tex]) = 5.12 °C/m
- Molar mass of diphenyl (C₁₂H₁₀) = 154.21 g/mol
### Step 2: Calculate the molality of the solution
The freezing point depression ([tex]\(\Delta T\)[/tex]) is related to the molality (m) of the solution by the equation:
[tex]\[ \Delta T = K_f \cdot m \][/tex]
Rearranging the equation for molality (m):
[tex]\[ m = \frac{\Delta T}{K_f} \][/tex]
Substituting the known values:
[tex]\[ m = \frac{2.00 \, °C}{5.12 \, °C/m} \][/tex]
[tex]\[ m = 0.390625 \, m \][/tex]
This means the molality of the solution is 0.390625 mol/kg.
### Step 3: Calculate the moles of diphenyl required
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We can use this relationship to find the number of moles of diphenyl that need to be dissolved:
[tex]\[ m = \frac{\text{moles of diphenyl}}{ \text{mass of benzene in kg}} \][/tex]
We need to convert the mass of benzene into kilograms:
[tex]\[ \text{mass of benzene} = 50.0 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.0500 \, \text{kg} \][/tex]
Now, solving for the moles of diphenyl:
[tex]\[ \text{moles of diphenyl} = m \times \text{mass of benzene in kg} \][/tex]
[tex]\[ \text{moles of diphenyl} = 0.390625 \, m \times 0.0500 \, \text{kg} \][/tex]
[tex]\[ \text{moles of diphenyl} = 0.01953125 \, \text{mol} \][/tex]
### Step 4: Calculate the mass of diphenyl needed
Finally, we use the molar mass of diphenyl to convert the moles of diphenyl to grams:
[tex]\[ \text{mass of diphenyl} = \text{moles of diphenyl} \times \text{molar mass of diphenyl} \][/tex]
[tex]\[ \text{mass of diphenyl} = 0.01953125 \, \text{mol} \times 154.21 \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of diphenyl} = 3.0119140625 \, \text{g} \][/tex]
### Conclusion
To lower the freezing point of 50.0 grams of benzene by 2.00 °C, you must dissolve approximately 3.01 grams of diphenyl (C₁₂H₁₀) in it.
