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Question
A solution containing 12.5 g of a non-electrolyte solute in 175 g of water gave boiling point of 100.70°C. Calculate the molecular mass of the solute.
(Given: Kb for water = 0.52 K kg mol−1)
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Solution
Given: Mass of solute = 12.5 g
Mass of solvent (water) = 175 g = 0.175 kg
Boiling point of solution = 100.70°C
Boiling point of pure water = 100°C
Molal boiling point constant (Kb) for water = 0.52 K kg mol−1
First, we will calculate the boiling point elevation (ΔTb),
Tb = Boiling point of solution − Boiling point of pure solvent
= 100.70 − 100.00
= 0.70 K
Now, we will calculate molality,
m = `"Moles of solute"/"Mass of solvent in kg"`
From the boiling point elevation formula:
m = `(DeltaT_b)/K_b`
= `0.70/0.52`
= 1.346 mol/kg
Now, we will calculate the moles of solute,
Moles of solute = m × Mass of solvent in kg
= 1.346 × 0.175
= 0.2355 moles
Now, the molecular mass of the solute will be,
Molecular mass = `"Mass of solute"/"Moles of solute"`
= `12.5/0.2355`
= 53.06 g/mol
Thus, the molecular mass of the solute is 53.06 g/mol.
