10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of O 2 at NTP. The remaining gas occupied 70 ml at NTP. This volume become 50 ml on treatment with KOH solution. The molecular formula of the hydrocarbon is

Question

10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of O 2 at NTP. The remaining gas occupied 70 ml at NTP. This volume become 50 ml on treatment with KOH solution. The molecular formula of the hydrocarbon is, C 2 H 4, C H 4, C 2 H 2, C 2 H 6

MARKS App · Accepted Answer

Suppose the molecular formula of hydrocarbon = C x H y C x H y g + x + y 4 O 2 g → xCO 2 g + y 2 H 2 O l Volume of hydrocarbon = 10 ml Decreases in volume = 10 + 80 – 70 = 20 ml Volume of C O 2 = 70 - 50 = 20 ml x = 20 10 = 2 V O 2 left = 50 ml V O 2 Consumed = 80 − 50 = 30 ml V O 2 Consumed = V H .C . × x + y 4 30 = 10 × 2 + y 4 y = 3 − 2 × 4 = 4 Molecular formula is C 2 H 4 .

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