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A certain substance X condenses at a temperature of 123.3 degree C . But if a 650. g sample of X is prepared with 24.6 g of urea ((NH2)2 CO) dissolved in it, the sample is found to have a condensation point of 124.3 degree C instead. Calculate the molal boiling point elevation constant Kb of X.Round your answer to 2 significant digits.

Respuesta :

Answer:

The molal boiling point elevation constant is 1.59 ≈  1.6 [tex]Kkgmol^{-1}[/tex]

Explanation:

To solve this question , we will make use of the equation ,

Δ[tex]T_{b} = i*K_{b} *m[/tex]

where ,

  • Δ[tex]T_{b}[/tex] is the change in boiling point of the substance [tex]X[/tex] ( °[tex]C[/tex] or [tex]K[/tex])
  • [tex]i[/tex] is the Vant Hoff Factor which = 1 in this case ( no unit )
  • [tex]K_{b}[/tex] is the mola boiling point elevation constant of X ( [tex]Kkgmol^{-1}[/tex])
  • [tex]m[/tex] is the molality of the solution which has [tex](NH_{2})_{2} CO[/tex] as the solute and  [tex]X[/tex] as the solution ([tex]molkg^{-1}[/tex])

  1. Δ[tex]T_{b}[/tex] = [tex]124.3 -123.3 = 1[/tex] °[tex]C[/tex] or [tex]K[/tex];
  2. [tex]i[/tex]=1;
  3. [tex]m[/tex]= [tex]\frac{moles of solute}{weight of solvent(kg)}[/tex][tex]molkg^{-1}[/tex]

           ∴ [tex]m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]

  • as the weight of [tex](NH_{2})_{2} CO[/tex] is [tex]60g[/tex] and thus number of moles = [tex]\frac{24.6}{60}[/tex]
  • and the weight of solvent in [tex]kg[/tex] is [tex]\frac{650}{1000}[/tex]

    4. [tex]K_{b}[/tex] ⇒ ?

∴

[tex]1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]

⇒ [tex]K_{b}[/tex] = [tex]1.59[/tex] ≈ 1.6 [tex]Kkgmol^{-1}[/tex]

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