A student dissolves 0.550 mol of a nonvolatile, nonelectrolyte solute in one kilogram of benzene (C6H6). What is the boiling point elevation of the resulting solution?
in degrees celsius
m = moles solute/kg solvent. Solve for m and substitute in the following:
delta T = Kb*m
Then add delta T to the normal boiling point.
To calculate the boiling point elevation of the resulting solution, you can use the formula:
ΔTb = Kb * m
Where:
ΔTb = boiling point elevation
Kb = molal boiling point elevation constant of the solvent
m = molality of the solution
In this case, the solvent is benzene (C6H6), and the given information is that 0.550 mol of a solute is dissolved in one kilogram of benzene.
First, you need to determine the molality (m) of the solution, which is defined as the number of moles of solute per kilogram of solvent.
m = moles of solute / mass of solvent (in kg)
m = 0.550 mol / 1 kg
m = 0.550 mol/kg
Next, you need to find the molal boiling point elevation constant (Kb) for benzene. The Kb value for benzene is 2.53 °C/m.
Now, substitute the known values into the formula:
ΔTb = 2.53 °C/m * 0.550 mol/kg
ΔTb = 1.39 °C
Therefore, the boiling point elevation of the resulting solution is 1.39 degrees Celsius.
To find the boiling point elevation of a solution, we can use the equation:
ΔTb = Kb * m * i
Where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point constant,
m is the molality of the solution, and
i is the van 't Hoff factor.
First, let's find the molality (m) of the solution. Molality is defined as the number of moles of solute (in this case, the nonvolatile solvent) per kilogram of solvent.
Given:
Number of moles of solute (nonvolatile, nonelectrolyte) = 0.550 mol
Mass of solvent (benzene) = 1 kg
Molality (m) = moles of solute / mass of solvent
= 0.550 mol / 1 kg
= 0.550 mol/kg
Next, we need the van 't Hoff factor (i). The van 't Hoff factor represents the number of particles into which a solute dissociates or ionizes in a solution. Since the given solute is a nonvolatile, nonelectrolyte, it does not dissociate or ionize, so the van 't Hoff factor (i) is 1.
Now, we need the molal boiling point constant (Kb) for benzene. The molal boiling point constant is specific to each solvent. For benzene, the molal boiling point constant (Kb) is 2.53 °C/m.
Plugging in the values we have:
Kb = 2.53 °C/m
m = 0.550 mol/kg
i = 1
ΔTb = (2.53 °C/m) * (0.550 mol/kg) * (1)
Calculating the boiling point elevation:
ΔTb = 1.3915 °C
Therefore, the boiling point elevation of the resulting solution is approximately 1.3915 °C.