A 0.85-g sample is dissolved in 0.150 kg of bromobenzene. Determine the molar mass of the solute if the solution boils at 429.0 K at 1 atm pressure. The normal boiling point of bromobenzene is 428.1 K and the ebullioscopic constant is 6.26 K kg mol-1.
A 0.85-g sample is dissolved in 0.150 kg of bromobenzene. Determine the molar mass of the solute if the solution boils at 429.0 K at 1 atm pressure. The normal boiling point of bromobenzene is 428.1 K and the ebullioscopic constant is 6.26 K kg mol-1.
deltaT=6.26(.85/molmass)/.240
detat T= 429.0 K -428.1 K
solve for molmass
To determine the molar mass of the solute in this problem, we can use the formula for boiling point elevation. The equation is:
deltaT = K_b * m * molality
Where:
- deltaT is the change in boiling point of the solvent
- K_b is the ebullioscopic constant of the solvent
- m is the mass of the solute in the solution
- molality is the molality of the solute in the solution
First, let's calculate the change in boiling point (deltaT):
deltaT = boiling point of the solution - normal boiling point of the solvent
deltaT = 429.0 K - 428.1 K
deltaT = 0.9 K
Next, let's calculate the molality of the solute in the solution:
molality = moles of solute / mass of solvent (in kg)
Given that the mass of the solute is 0.85 g and the mass of the solvent is 0.150 kg, we can convert the mass of the solute to moles:
moles of solute = mass of solute / molar mass of solute
moles of solute = 0.85 g / molar mass of solute
Substituting this into the molality equation:
molality = (0.85 g / molar mass of solute) / 0.150 kg
Now, we can substitute the known values into the boiling point elevation equation:
0.9 K = (6.26 K kg mol^-1) * (0.85 g / molar mass of solute) / 0.150 kg
To solve for the molar mass of the solute, we need to rearrange the equation and isolate it:
molar mass of solute = (6.26 K kg mol^-1) * (0.85 g / (0.150 kg * 0.9 K))
Now, we can calculate the molar mass of the solute by substituting the known values:
molar mass of solute = (6.26 K kg mol^-1) * (0.85 g / (0.150 kg * 0.9 K))