The degree of dissociation of XY in case of following reaction is found to be 0.4 from vapour density measurement. If the observed vapour density of X Y be 100 then its molecular weight is XY→ X+Y

Question

The degree of dissociation of XY in case of following reaction is found to be 0.4 from vapour density measurement. If the observed vapour density of X Y be 100 then its molecular weight is XY→ X+Y  (A)  140  (B)  70  (C)  280  (D)  210

Tardigrade Admin · Accepted Answer

The degree of dissociation (α)=(DT-D0/D0) 0.4=(DT-100/100) DT=40+100=140 ∴ DT=( text mol. wt. /2) ∵ Mol. wt. =DT × 2=140 × 2 =280

Q. The degree of dissociation of $X Y$ in case of following reaction is found to be $0.4$ from vapour density measurement. If the observed vapour density of $X Y$ be $100$ then its molecular weight is $X Y \to X + Y$

$140$

$70$

$280$

$210$

The degree of dissociation $(α) = \frac{D _{T} - D _{0}}{D _{0}}$
$0.4 = \frac{D _{T} - 100}{100}$
$D_{T} = 40 + 100 = 140$
$∴ D_{T} = \frac{mol. wt.}{2}$
$∵$ Mol. wt. $= D_{T} \times 2 = 140 \times 2$
$= 280$

Q. The degree of dissociation of XY in case of following reaction is found to be 0.4 from vapour density measurement. If the observed vapour density of XY be 100 then its molecular weight is XY→X+Y

140

70

280

210