Question

Boiling point of 0.01M AB2 which is 10% dissociated in aqueous medium (kbH2o=0.52) A{+} &{ b-}?

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The boiling point of  AB₂ in aqueous solution will be 373.006K.

Explanation:

The dissociation of AB₂ in aq. solution as:

         AB₂  →  A²⁺  +  2B⁻

The total number of ions $n=3$

Given 10% dissociation means $\alpha =0.1$

Van't Hoff factor   $i=1+(n-1)\alpha$

       $i=1+(3-1)(0.1)\\i=1.2$

Now 1000ml solution of  AB₂ $=0.01 mol$

because the density of water $=1gml^{-1}$

1000g of solvent  of  AB₂ $=0.01 mol$       ( neglect mass of AB₂ )                   Molality $m=\frac{0.01}{1000}*1000=0.01m$

We know that Δ$T_{b}$ $=i*m*K_{b}$

Put the value of  $K_{b}=0.52$, m and $i$ in above equation:

       Δ$T_{b}=(1.2)(0.01)(0.52)=0.006$

Now  Δ$T_{b}$ = Δ$T_{b}(solution)$ - Δ$T_{b}(solvent)$

      $0.006=$ Δ $T_{b}(solution)-373$

     Δ$T_{b}(solution)$ $=373+0.006=373.006K$

Therefore the boiling point of the aqueous solution of AB₂ will be 373.006K.