what is the number of moles of nonvolatile nonelectrolyte dissolved in 3.5 kg of water when the solution boils at 101.86 degrees C
delta T = Kb*m
delta T = 101.86 - normal boiling point.
Substitute and solve for m = molality.
m = mols/Kg solvent. You know kg solvent and m, solve for mols.
To find the number of moles of a nonvolatile nonelectrolyte dissolved in water, we can use the equation:
ΔTb = Kb * m
where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point elevation constant, and
m is the molality of the solution.
First, we need to calculate the boiling point elevation (ΔTb). The boiling point elevation is the difference between the boiling point of the solvent (in this case, water) and the boiling point of the solution. The boiling point of pure water is 100°C, and the boiling point of the solution is given as 101.86°C. Therefore, the ΔTb is:
ΔTb = 101.86°C - 100°C
ΔTb = 1.86°C
Next, we need to determine the molal boiling point elevation constant (Kb) for water. The Kb value for water is 0.512 °C/m.
Now, we can rearrange the equation to solve for m:
m = ΔTb / Kb
m = 1.86°C / 0.512 °C/m
m ≈ 3.63 mol/kg
Since we know the mass of water is 3.5 kg, we can calculate the number of moles using the molality:
moles = m * mass of solvent
moles = 3.63 mol/kg * 3.5 kg
moles ≈ 12.71 moles
Therefore, there are approximately 12.71 moles of the nonvolatile nonelectrolyte dissolved in 3.5 kg of water when the solution boils at 101.86 degrees C.