To calculate the molar mass of the solute in the first question, you need to use the formula:
∆T = Kf * m * i
Where:
∆T = change in freezing point (in this case, -1.454°C)
Kf = freezing point depression constant of the solvent (7.27°C/m)
m = molality of the solution (mol solute / kg solvent)
i = van't Hoff factor (the number of particles into which the solute dissociates, often assumed to be 1 for non-ionic solutes)
First, we need to calculate 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)
Mass of solvent = 50.0 g = 0.050 kg
Mass of solute = 1.33 g = 0.00133 kg
So, the molality (m) = 0.00133 kg solute / 0.050 kg solvent = 0.0266 mol/kg
Now, let's substitute the given values into the freezing point depression equation:
-1.454°C = (7.27°C/m) * (0.0266 mol/kg) * i
We can now solve for i:
i = -1.454°C / [(7.27°C/m) * (0.0266 mol/kg)] = -1.454°C / (0.194° molality) ≈ -7.49
Since i cannot be negative and generally assumes to be positive for non-ionic solutes, we take the absolute value of i:
i ≈ 7.49
Now, we can use the molar mass formula to calculate the molar mass of the solute:
molar mass (g/mol) = mass (g) / moles
mass = 1.33 g
moles = mass (g) / molar mass (g/mol) = 1.33 g / molar mass (g/mol)
Substituting back into the equation, we get:
-7.49 = (7.27°C/m) * (0.00133 kg) * (1 / molar mass (g/mol))
Now, solve for the molar mass:
molar mass (g/mol) = 7.27°C / (0.00133 kg * 7.49) ≈ 7.27°C / 0.0099667 kg ≈ 729.1 g/mol
Therefore, the molar mass of the solute is approximately 729.1 g/mol.
Now, moving on to the second question about the normal boiling point of an aqueous solution. To calculate the boiling point elevation, we use a similar formula:
∆T = Kb * m * i
Where:
∆T = change in boiling point (unknown)
Kb = boiling point elevation constant of the solvent (0.51°C/m)
m = molality of the solution (unknown)
i = van't Hoff factor (assumed to be 1 for most aqueous solutions)
To find the boiling point elevation (∆T), we rearrange the equation:
∆T = Kb * m * i
Given:
∆T = -1.04°C (because it is a freezing point, it is negative, but boilings points are positive)
Kb = 0.51 (K·kg)/mol (boiling point elevation constant for water)
i = 1 (assuming a non-ionic solute)
Now, substitute the given values into the equation:
-1.04°C = (0.51 (K·kg)/mol) * m * 1
Rearrange the equation to solve for m:
m = -1.04°C / (0.51 (K·kg)/mol) ≈ -2.04 mol/kg
The molality (m) cannot be negative, so we'll take the absolute value:
m ≈ 2.04 mol/kg
We can now substitute the value of m into the boiling point elevation equation to solve for the change in boiling point (∆T):
∆T = (0.51 (K·kg)/mol) * (2.04 mol/kg) * 1 ≈ 2.06°C
To find the normal boiling point, we add this change in boiling point (∆T) to the boiling point of pure water, which is 100.0°C:
Normal boiling point = 100.0°C + 2.06°C ≈ 102.06°C
Therefore, the normal boiling point of the aqueous solution is approximately 102.06°C.