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40 ml of a hydrocarbon undergoes combustion in 260ml of oxygen and gives 160ml of CO_2.If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is

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C₄H₈

C₆H₁₄

C₃H₈

C₄H₁₀

answer is D.

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Detailed Solution

\large {C_x}{H_y} + \left( {x + \frac{y}{4}} \right){O_2} \to xC{O_2} + \frac{y}{2}{H_2}O\

40\;ml\;\;\;\;\;\large 40\left( {x + \frac{y}{4}} \right)\;\;\;\;\;\;\;\;\;\;\;\40x\

Given

40x = 160

x = 4

\large 40\left( {x + \frac{y}{4}} \right)\ = 260

4x + y = 26

4(4) + y = 26

y = 10

Hydrocarbon is C₄H₁₀

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