contestada

Given that the vapor pressure of water is 17.54 Torr at 20 ∘ C, calculate the vapor‑pressure lowering, Δ P , of an aqueous solution that is 2.50 m in sucrose ( C 12 H 22 O 11 ) ., C12H22O11, and (b) aluminum chloride. Assume 100% dissociation for electrolytes

(a) sucrose C12H22011 (delta P) = ?
(b) aluminum chloride (delta P) = ?

Respuesta :

Answer:

A) The lowering in vapour pressure is 0.7839 torr

B) The lowering in vapour pressure is 3.1572 torr

Explanation:

The relative lowering of vapour pressure formula is ::

  • Δ[tex]\frac{P}{P_{0}} = i*\frac{m*M}{1000}[/tex]

where ,

  • Δ[tex]P[/tex] is the lowering of vapour pressure ([tex]atm/torr[/tex])
  • [tex]P_{0} is the vapour pressure of the solvent ([tex]atm/torr[/tex])
  • [tex]m[/tex] is the molality of the solution ( [tex]molkg^{-1}[/tex] )
  • [tex]M[/tex] is the molecular weight of the solvent ( [tex]g[/tex] )
  • [tex]i[/tex] is the vant hoff factor or dissociation constan

The values :

A)

  • [tex]P_{0}=17.54torr[/tex]
  • [tex]m=2.50m[/tex]
  • [tex]i=1[/tex]
  • [tex]M(H_2O)=2+16=18g[/tex]

⇒ΔP = [tex]17.54*(1)*\frac{2.5*18}{1000}\\=0.7893torr[/tex]

B)

  • [tex]P_{0}=17.54torr[/tex]
  • [tex]m=2.50m[/tex]
  • [tex]i=4[/tex] ( since, it's [tex]AlCl_3[/tex] it spilts into 1 [tex]Al^{3+}[/tex] and 3 [tex]Cl^-[/tex] ions and as a results, there are 4 ions.
  • [tex]M(H_2O)=2+16=18g[/tex]

⇒ΔP = [tex]17.54*4*\frac{2.50*18}{1000} \\=3.1572torr[/tex]

Otras preguntas