Question

The boiling point of benzene is 353.23 k. When 1.80 g of a non volatile non dissociative solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 k. If the kb for benzene is 2.64 k kg mol1, then find the molar mass of the solute.

Answer 1

Answer :

• Molar mass of solute = 58 g/mol.

Step-by-step explanation:

Given that,

• Mass of solvent, w₁ = 90 g.
• Mass of solute, w₂ = 1.80 g.
• Elevation in boiling point, $\rm{{\Delta}T_b}$ = 354.11 - 353.23 K = 0.88 K.
• Molal elevation constant, $\rm{K_b}$ = 2.53 K kg mol⁻¹.

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The relation between elevation of boiling point and molecular mass is:

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

Substituing the given values in the formula, we get,

$\implies\rm{M_2=\dfrac{1000\:g\:kg^{-1}\times{}2.53\:K\:kg\:mol^{-1}\times{}1.80\:g}{90\:g\times{}0.88\:K}}$

$\implies\rm{M_2=\dfrac{253\times{}18}{90\times{}0.88}\:g\:mol^{-1}}$

$\implies\rm{M_2=\dfrac{4554}{79.2}\:g\:mol^{-1}}$

$\implies\rm{M_2=57.5\approx}\:\bold{58\:g\:mol^{-1}.}$