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Q.

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

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a

46%

b

68%

c

6.8%

d

60%

answer is B.

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Detailed Solution

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Repeat

\large \alpha = \frac{{D - d}}{{d\left( {n - 1} \right)}}
\large PC{l_{5\left( g \right)\,}}\, \rightleftharpoons \,PC{l_{3\left( g \right)}}\, + \,C{l_{2\left( g \right)}}

n = 2

D = 104.16

d = 62

\large \alpha = \frac{{D - d}}{{d\left( {n - 1} \right)}}
\large \alpha = \frac{{\left( {104.16 - 62} \right)}}{{62\left( {2 - 1} \right)}}
\large \alpha = \frac{{42.16}}{{62}}
\large \boxed{\alpha = 0.68}

Percentage dissociation of PCl5 = 0.68×100 =68%

 

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