A solution is made by dissolving 3.5 moles of sodium chloride (NaCl) in 155 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution?
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molality = moles/kg solvent
Solve for molality
delta T = i*Kb*molality
Solve for delta T, then boiling point knowing that the normal boiling point for water is 100 C.
i is the van't Hoff factor, which for NaCl = 2.
i was 1 in the previous post.
To find the boiling point of the solution, we need to use the equation:
ΔTb = Kb * m
Where:
ΔTb = change in boiling point
Kb = molal boiling point constant
m = molality of the solution
First, we need to calculate the molality (m) of the solution. Molality is the number of moles of solute (in this case, NaCl) divided by the mass of the solvent (water) in kilograms.
Given:
Number of moles of NaCl (solute) = 3.5 moles
Mass of water (solvent) = 155 grams
To convert grams to kilograms, divide by 1000:
Mass of water (solvent) = 155 grams / 1000 = 0.155 kg
Now, we can calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
m = 3.5 moles / 0.155 kg
m ≈ 22.58 mol/kg (rounded to two decimal places)
Next, we'll use the molal boiling point constant (Kb) and the calculated molality (m) to find the change in boiling point (ΔTb):
ΔTb = Kb * m
ΔTb = 0.51 °C/m * 22.58 mol/kg
Calculating ΔTb:
ΔTb ≈ 11.53 °C
Finally, we need to add the change in boiling point (ΔTb) to the normal boiling point of water (100 °C):
Boiling point of the solution = 100 °C + 11.53 °C
Boiling point of the solution ≈ 111.53 °C
Therefore, the boiling point of the solution is approximately 111.53 °C.