What is the boiling point of a solution that contains 1.20 mole of sodium chloride in 1000 g water? KbH2O = 0.512C/ molal
To find the boiling point of a solution, we can use the formula:
ΔTb = Kb * m
where ΔTb is the boiling point elevation, Kb is the molal boiling point constant, and m is the molality of the solution.
First, we need to calculate the molality of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Given that the solution contains 1.20 moles of sodium chloride (NaCl) and 1000 g of water, we need to convert the mass of water to kilograms:
Mass of water = 1000 g = 1000 / 1000 kg = 1 kg
Now, we can calculate the molality:
m = 1.20 moles / 1 kg = 1.20 mol/kg
Next, we can calculate the boiling point elevation (ΔTb) using the molal boiling point constant (Kb), which is given as 0.512 °C/m:
ΔTb = Kb * m
= 0.512 °C/m * 1.20 mol/kg
= 0.6144 °C
Finally, to find the boiling point of the solution, we add the boiling point elevation to the boiling point of pure water, which is 100 °C:
Boiling point of solution = boiling point of water + ΔTb
= 100 °C + 0.6144 °C
≈ 100.6144 °C
Therefore, the boiling point of the solution that contains 1.20 moles of sodium chloride in 1000 g of water is approximately 100.6144 °C.