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Question
The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute.
[Kb for benzene = 2.53 K kg mol-1]
Solution
Given:
Kb for benzene = 2.53 K kg mol–1
Mass of solute = W2 = 1.8 g = 1.8 x 10–3 kg
Mass of solvent = W1 = 90 g = 90 x 10–3 kg
Boiling point of solution = Tb = 354.11 K
Boiling point of pure solvent = `T_b^'` = 353.23 K
To Find: Molar mass of solute (M2)
`Delta T_b=(K_bxxW_2)/(M_2xxW_1)`
→ Elevation in boiling point (`Delta`Tb) = `T_b - T_b^'`
`Delta T_b=354.11-353.23`
`Delta T_b=0.88K`
From Formula
`M_2=(K_bxxW_2)/(DeltaT_bxxW_1)`
`M_2=(2.53xx1.8xx10^-3)/(0.88xx90xx10^-3)=4.554/79.2`
=0.0575 kg `mol^-1`
=57.5 g `mol^-1`
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