Boiling point of 0.01 M AB 2 which is 10 % dissociated in aqueous medium ( K b H2O =0.52) as A +2 and B -

Question

Boiling point of 0.01 M AB 2 which is 10 % dissociated in aqueous medium ( K b H2O =0.52) as A +2 and B - (A) 273.006 K (B) 373.006 K (C) 0.006 K (D) 272.006 K

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/5f148337f9b6edc3b6fcfaf468f65d4f-.png" /> Δ T b = i ( K b × m ) T b - T b °= i ( K b × m ) i =(1-α+α+2 α/1)=1+2 α T b -373=1.2[0.52 × 0.01] =1+(2 × 10/100) Tb-373=0.00624 i =⇒ 1.2 T b =373.006 K

Q. Boiling point of $0.01 M A B_{2}$ which is $10%$ dissociated in aqueous medium $(K_{b_{H_{2} o}} = 0.52)$ as $A^{+ 2}$ and $B^{-}$

273.006 K

373.006 K

0.006 K

272.006 K

$Δ T_{b} = i (K_{b} \times m)$
$T_{b} - T_{b}^{\circ} = i (K_{b} \times m)$
$i = \frac{1 - α + α + 2 α}{1} = 1 + 2 α$
$T_{b} - 373 = 1.2 [0.52 \times 0.01]$
$= 1 + \frac{2 \times 10}{100}$
$T_{b} - 373 = 0.00624$
$i =\Rightarrow 1.2$
$T_{b} = 373.006 K$

Q. Boiling point of 0.01MAB2​ which is 10% dissociated in aqueous medium (KbH2​o​​=0.52) as A+2 and B−