At `25^(@)C, K_(c )` for the reaction `3C_(2)H_(2(g))hArr C_(6)H_(6(g))` is 4.0. If the equilibrium concentration of `C_(2)H_(2)` is 0.5 mol. `lit^(-1)`. What is the concentration of `C_(6)H_(6)`?
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The correct Answer is:
`[C_(6)H_(6)]="0.5 mol. Lit"^(-1)`
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