91Ó°ÊÓ

The gaseous hydrocarbon reacts with oxygen to form carbon dioxide (COâ‚‚) and water (Hâ‚‚O). The general equation for the reaction is: \[C_xH_y + \left(\frac{x}{2} + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\] The observed contraction after the explosion is 2.5V. In terms of volume, 1V of the original mixture (hydrocarbon + oxygen) has become 1 - 2.5 = -1.5V after the explosion. This indicates that 1.5V of new gaseous products are formed.

Initially, after the explosion, the contraction is given as 2.5V. After treatment with potash, there is a further contraction of 2V. Potash is a strong base and reacts only with carbon dioxide, forming water and potassium carbonate. The equation for the reaction is: \[2KOH + CO_2 \rightarrow K_2CO_3 + H_2O\] Since the further contraction after treatment with potash is 2V, and potash reacts with COâ‚‚, it implies that there were 2V of COâ‚‚ formed after the explosion.

We know that the hydrocarbon contained x moles of carbon and y moles of hydrogen. The volume of COâ‚‚ formed is equal to the volume of carbon present in the hydrocarbon. We have the following relationship between volumes: \[V_\text{hydrocarbon} = V_C = xV \quad \text{and} \quad V_\text{oxygen} = \frac{x}{2}V + \frac{y}{4}V\] where \(V_C\) is the volume of carbon in the hydrocarbon. Since \(V_\text{oxygen} = (1 - 1.5) V\), we can write: \[xV = \frac{x}{2}V + \frac{y}{4}V \Rightarrow x = \frac{y}{2}\] The volume of carbon dioxide formed is 2V, which means: \[2V = xV \Rightarrow x = 2\] Substituting the value for x, we get: \[2 = \frac{y}{2} \Rightarrow y = 4\] Thus, the molecular formula of the hydrocarbon is: \[C_2H_4\] Therefore, the correct answer is (d) \(C_2H_4\).