The boiling point elevationof an aqueous sucrose solution is found to be 0.39 degrees Celsius. What mass of sucrose (molar mass=342.30g/mol) is contained in 500g of the solution? Kb (water)=0.512degrees C/m
delta T = Kb*molality
Solve for m
m = moles solute/kg solvent
Before going further, are you sure you have copied the problem correctly? There is a difference between 500 g SOLUTION and 500 g SOLVENT.
Also, if it is 500 g solution, a density is needed.
To solve this problem, we can use the formula for boiling point elevation:
ΔTb = Kb * m * i
Where:
- ΔTb is the boiling point elevation,
- Kb is the molal boiling point constant of the solvent (in this case, water),
- m is the molality, which is the number of moles of solute per kilogram of solvent, and
- i is the van't Hoff factor, which represents the number of particles into which the solute dissociates.
In this case, we need to find the mass of sucrose, so we first need to calculate the molality (m) of the solution.
m = n solute / mass solvent
Given that the mass of the solution is 500g and the molar mass of sucrose is 342.30g/mol, we can calculate the number of moles of sucrose in the solution (n solute):
n solute = mass solute / molar mass
n solute = 500g / 342.30g/mol
Now, let's substitute the values into the equation for boiling point elevation:
ΔTb = Kb * m * i
We already know that Kb (water) = 0.512°C/m. However, we need to determine the van't Hoff factor (i) for sucrose in water. Since sucrose does not dissociate into ions, i = 1.
Substituting the values:
0.39°C = 0.512°C/m * (n solute / 0.5kg) * 1
Now, let's rearrange the equation to solve for n solute:
n solute = (0.39°C / 0.512°C/m) * 0.5kg
Finally, we can calculate the mass of sucrose:
mass solute = n solute * molar mass
Substituting the values into the equation:
mass solute = (n solute) * (342.30g/mol)