Answer
7.0 M
Work Step by Step
$K_{eq}$ = $\frac{[SbCl_{3}][Cl_{2}]}{[SbCl_{5}]}$
From the above equation, we get
$[SbCl_{5}] = \frac{[SbCl_{3}][Cl_{2}]}{K_{eq}}$ = $\frac{0.0255\times0.135}{4.9\times10^{-4}}$ = 7.0 M ( at $248^{\circ}$ C)
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