CBSE Class 12 · 2010 · Delhi · SolutionsShort Answer

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A solution prepared by dissolving 1.25 g of oil of wintergreen (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10°C and K

_{b}

for benzene = 2.53°C kg mol

^{- 1}

)

Answer: M ≈ 152 g mol⁻¹

Solution

Elevation:

Δ T_{b} = 80.31 - 80.10 = 0.21 K

Molality:

m = \frac{0.21}{2.53} = 0.0830 mol kg^{- 1}

Moles in 99.0 g benzene =

0.0830 \times 0.099 = 8.217 \times 1 0^{- 3}

mol.
Molar mass:

M = \frac{1.25}{8.217 \times 1 0 ^{- 3}}

M = 152.1 g mol^{- 1}

CBSE Class 12 Board · 2010 · Delhi · Subjective

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