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Standard X

Chemistry

Relationship between Vapor Density & Relative Molecular Mass

An unknown co...

Question

An unknown compound $A$ dissociates at $500^{o} C$ to give products as follows-
$A (g) ⇌ B (g) + C (g) + D (g)$
Vapour density of the equilibrium mixture is $50$ when it dissociates to the extent to $10 %$ . What will be the molecular weight of compound $A$ ?

120

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130

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134

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140

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Solution

The correct option is A $120$
$α = \frac{D - d}{(n - 1) d}$
Where $α =$ Degree of dissociation $= 10$ % $= 0.1$
$D =$ Initial vapor density
$d =$ Vapor density at equilibrium $= 50$
$n =$ Moles of gaseous product $= 3$
$0.10 = \frac{D - 50}{(3 - 1) 50}$
$D = 60$
Vapor density $= 60$
Molecular mass $= 2 \times v a p o r d e n s i t y$
$= 2 \times 60 = 120 g m o l^{- 1}$

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An unknown compound A dissociates at 500oC to give products as follows-A(g)⇌B(g)+C(g)+D(g)Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10%. What will be the molecular weight of compound A?

The correct option is A 120α=D−d(n−1)dWhere α=Degree of dissociation=10%=0.1D=Initial vapor densityd=Vapor density at equilibrium=50n=Moles of gaseous product=30.10=D−50(3−1)50D=60 Vapor density=60Molecular mass=2×vapor density =2×60=120gmol−1

An unknown compound $A$ dissociates at $500^{o} C$ to give products as follows-
$A (g) ⇌ B (g) + C (g) + D (g)$
Vapour density of the equilibrium mixture is $50$ when it dissociates to the extent to $10 %$ . What will be the molecular weight of compound $A$ ?