To calculate the boiling point elevation, we'll need to use the equation:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation
Kb is the molal boiling point elevation constant (2.53 °C/m for benzene)
m is the molality of the solution, which can be calculated using the mass of biphenyl and the molar mass of benzene.
First, let's find the number of moles of biphenyl:
moles biphenyl = mass biphenyl / molar mass biphenyl
moles biphenyl = 0.0485 kg / 154.21 g/mol = 0.0003143 mol
Next, let's calculate the mass of benzene in the solution:
mass benzene = volume solution * density benzene
mass benzene = 350.0 mL * 0.877 g/mL = 307.45 g
Then, calculate the number of moles of benzene:
moles benzene = mass benzene / molar mass benzene
moles benzene = 307.45 g / 78.11 g/mol = 3.9381 mol
Now, let's calculate the molality of the solution:
molality = moles solute / kg solvent
molality = 0.0003143 mol / 0.30745 kg = 0.00102 m
Finally, calculate the boiling point elevation:
ΔTb = 2.53 °C/m * 0.00102 m = 0.0025866 °C
To find the boiling point of the solution, add the boiling point elevation to the boiling point of pure benzene:
boiling point of solution = boiling point of pure benzene + ΔTb
boiling point of solution = 80.1 °C + 0.0025866 °C = 80.1026 °C
Therefore, the boiling point of this solution would be approximately 80.1 °C.
To calculate the vapor pressure of the solution, we'll need to use Raoult's Law:
P solution = X benzene * P benzene
Where:
P solution is the vapor pressure of the solution
X benzene is the mole fraction of benzene in the solution
P benzene is the vapor pressure of pure benzene.
The mole fraction of benzene can be calculated using the moles of benzene and biphenyl:
X benzene = moles benzene / (moles benzene + moles biphenyl)
X benzene = 3.9381 mol / (3.9381 mol + 0.0003143 mol) = 0.99992
The vapor pressure of the solution can be calculated as:
P solution = X benzene * P benzene
P solution = 0.99992 * 24.4 kPa = 24.398 kPa
Therefore, the vapor pressure of the solution would be approximately 24.4 kPa.
To calculate the osmotic pressure, we'll need to use the equation:
π = MRT
Where:
π is the osmotic pressure in atm
M is the molarity of the solution, which can be calculated using the moles of biphenyl and the volume of the solution
R is the ideal gas constant (0.0821 L*atm/mol*K)
T is the temperature in Kelvin.
First, let's calculate the molarity of the solution:
molarity = moles biphenyl / volume solution
molarity = 0.0003143 mol / 0.350 L = 0.000897 mol/L
Next, let's convert the temperature to Kelvin:
T = 40.0 °C + 273.15 = 313.15 K
Now, let's calculate the osmotic pressure:
π = 0.000897 mol/L * 0.0821 L*atm/mol*K * 313.15 K
π = 0.0237 atm
Therefore, the osmotic pressure of this solution at 40.0 °C would be approximately 0.0237 atm.
14 0.0485 kg of biphenyl (C12H10) is dissolve in benzene (C6H6) to create a solution with a total volume of 350.0 mL. (Assume the change in volume is negligible) If the boiling point of pure benzene is 80.1 °C, then what would be the boiling point of this solution in °C? (Kb for benzene is 2.53 °C/m and the density of benzene is 0.877 g/mL) b > O of 2 points earned 10 attempts remaining If the vapor pressure of pure benzene is 24.4 kPa at 40.0 °C, then what will the vapor pressure of the solution be in kPa? (Consider biphenyl to be nonvolatile and the density of benzene is 0.877 g/mL) C > O of 2 points earned 10 attempts remaining What would be the osmotic pressure (in atm) of this solution at 40.0 °C? Assume the density of the solution is the same as benzene, 0.877 g/mL. d > 0 of 2 points earned 10 attempts remaining
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