Vapour density of PCl5 is 104.16 but when heated to 230° C its vapour density is reduced to 62. The degree of dissociation of PCl5 at this temperature will be

Question

Vapour density of PCl5 is 104.16 but when heated to 230° C its vapour density is reduced to 62. The degree of dissociation of PCl5 at this temperature will be (A) 6.8% (B) 68% (C) 46% (D) 64%

Tardigrade Admin · Accepted Answer

PCl5(g) leftharpoons PCl3 (g) + Cl2(g) Here y =2 x = (D -d/d(y-1)) = (104.16 - 62/62) = 0.68 = 68 %

Q. Vapour density of $PC l_{5}$ is $104.16$ but when heated to $23 0^{\circ} C$ its vapour density is reduced to 62. The degree of dissociation of $PC l_{5}$ at this temperature will be

6.8%

68%

46%

64%

$PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g)$
Here $y = 2$
$x = \frac{D - d}{d ( y - 1 )} = \frac{104.16 - 62}{62}$
$= 0.68 = 68%$

Q. Vapour density of PCl5​ is 104.16 but when heated to 230∘C its vapour density is reduced to 62. The degree of dissociation of PCl5​ at this temperature will be

6.8%

68%

46%

64%

PCl5​(g)⇌PCl3​(g)+Cl2​(g) Here y=2 x=d(y−1)D−d​=62104.16−62​ =0.68=68%

Q. Vapour density of $PC l_{5}$ is $104.16$ but when heated to $23 0^{\circ} C$ its vapour density is reduced to 62. The degree of dissociation of $PC l_{5}$ at this temperature will be

$PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g)$
Here $y = 2$
$x = \frac{D - d}{d ( y - 1 )} = \frac{104.16 - 62}{62}$
$= 0.68 = 68%$