Question

0.70 g of an organic compound when dissolved in 32g of acetone produces an elevation of 0.25oC0.25oC in the boiling point. Calculate the molecular mass of organic compound (Kb for acetone =1.72Kmol−1=1.72Kmol−1).​

Answer 1

Answer:

• Molecular mass of compound, M₂ = 150.5 g/mol.

Step-by-step explanation:

Given that,

• Mass of solvent, w₁ = 32 g.
• Mass of solute, w₂ = 0.70 g.
• Elevation of boiling point $\rm{({\Delta}T_b)}$ = 0.25°C.
• Molal elevation constant, $\rm{K_b}$ = 1.72 K kg mol⁻¹

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Using the relation,

$\implies\rm{M_2=\dfrac{1000\times{}K_b\times{}w_2}{w_1\times{}{\Delta}T_b}}$

We get,

$\implies\rm{M_2=\dfrac{1000\:g\:kg^{-1}\times{}1.72\:K\:kg\:mol^{-1}\times{}0.70\:g}{32\:g\times{}0.25\:K}}$

$\implies\rm{M_2=\dfrac{1204}{8}}$

$\implies\rm{M_2=}\:\bold{150.5\:g\:mol^{-1}.}$