Question

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AT ‘t°C’ TEMPERATURE, THE OBSERVED VAPOUR DENSITY OF ‘A’ IS ‘17.5’ FOR THE GASEOUS REACTION $\rm\red{2A\:\rightleftharpoons\:B\:+\:2C\:}$ . IF MOLECULAR WEIGHT OF ‘A’ IS “48”, THEN THE PERCENTAGE DISSOCIATION OF ‘A’ WILL BE :- $<font color=baby>$

[A] 70.27%
[B] 74.28%
[C] 37.14%
[D] 85.71% $<font color=red>$

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Answer 1

ANSWER:

• Percentage of dissociation of A = 74.28 %.

GIVEN:

• Vapour density of A = 17.5.

• Molecular weight of A = 48.

TO FIND:

• Percentage of dissociation of A.

EXPLANATION:

$\sf 2A_{(g)} \ \rightleftharpoons \ B_{(g)}+2C_{(g)}$

Let the percentage of dissociation of A = x %.

$\boxed{\bold{\large{\gray{x \% =\dfrac{n_1}{\Delta n} \left( \dfrac{D - d}{d} \right) \times 100}}}}$

$\sf d = Vapour \ density \ of \ A = 17.5$

$\boxed{\bold{\large{\gray{D = \dfrac{Molecular \ weight}{2}}}}}$

$\sf D = \dfrac{48}{2}$

$\sf D = 24$

$\sf n_1 = 2$

$\boxed{\bold{\large{\gray{\Delta n = (n_2 + n_3 ) - n_1}}}}$

$\sf n_1 = 2, \ n_2 = 1, \ n_3 = 2$

$\sf \Delta n = (2 + 1) - 2 = 1$

$\sf \dfrac{n_1}{\Delta n} = \dfrac{2}{1} = 2$

$\sf x \ \% = \dfrac{2(24 - 17.5)}{17.5} \times 100$

$\sf x \ \% = \dfrac{2(6.5)}{17.5} \times 100$

$\sf x \ \% = 2(0.3714) \times 100$

$\sf x \ \% =0.7428\times 100$

$\sf x \ \% =74.28\ \%$

HENCE PERCENTAGE OF DISSOCIATION = 74.28 % .