1.0 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of the solution is (Kb for H2O= 0.52 K kg/mol)

Question

1.0 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of the solution is (Kb for H2O= 0.52 K kg/mol) (A) 375.5 K (B) 374.04 K (C) 377.12 K (D) 373.25 K

Tardigrade Admin · Accepted Answer

underset1-αx3y2 leftharpoons undersetn α3x2++ undersetmα2y3- for complete ionization, i=1+(m+n-1) α i=1+(2+3-1)× 0.25=1+1=2 Δ Tb=i× Kb × m=2× 0.52× 1=1.04 B.P. of solution (Tb)=Δ Tb+T°b=1.04+373 =374.04 K

Q. $1.0$ molal aqueous solution of an electrolyte $X_{3} Y_{2}$ is 25% ionized. The boiling point of the solution is $(K_{b}$ for $H_{2} O = 0.52 K k g / m o l)$

$375.5 K$

$374.04 K$

$377.12 K$

$373.25 K$

$1 - α x_{3} y_{2} ⇌ n α 3 x^{2 +} + m α 2 y^{3 -}$ for complete ionization,
$i = 1 + (m + n - 1) α$
$i = 1 + (2 + 3 - 1) \times 0.25 = 1 + 1 = 2$
$Δ T_{b} = i \times K b \times m = 2 \times 0.52 \times 1 = 1.04$
B.P. of solution $(T_{b}) = Δ T_{b} + T_{b}^{\circ} = 1.04 + 373$
$= 374.04 K$

Q. 1.0 molal aqueous solution of an electrolyte X3​Y2​ is 25% ionized. The boiling point of the solution is (Kb​ for H2​O=0.52Kkg/mol)

375.5K

374.04K

377.12K

373.25K