7.5 ml of a gaseous hydrocarbon was exploded with 36 ml of oxygen. The volume of gases on cooling was found to be 28.5 ml , 15 ml of which was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol. If all volumes are measured under same conditions, the formula of hydrocarbon is

Question

7.5 ml of a gaseous hydrocarbon was exploded with 36 ml of oxygen. The volume of gases on cooling was found to be 28.5 ml , 15 ml of which was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol. If all volumes are measured under same conditions, the formula of hydrocarbon is (A) C 3 H 4 (B) C 2 H 4 (C) C 2 H 6 (D) C 3 H 6

Tardigrade Admin · Accepted Answer

underset7.5 mlCxHy + underset36 ml(X + (Y/4))O2 → XCO2 + (Y/2)H2O 36-7.5( x +( y /4))+7.5 x =28.5 36-7.5(15+( y /4))+7.5 x =28.5 y =4 x =2 So formula = C 2 H 4

$C_{3} H_{4}$

$C_{2} H_{4}$

$C_{2} H_{6}$

$C_{3} H_{6}$

$7.5 m l C_{x} H_{y} + 36 m l (X + \frac{Y}{4}) O_{2} \to XC O_{2} + \frac{Y}{2} H_{2} O$
$36 - 7.5 (x + \frac{y}{4}) + 7.5 x = 28.5$
$36 - 7.5 (15 + \frac{y}{4}) + 7.5 x = 28.5$
$y = 4$
$x = 2$
So formula $= C_{2} H_{4}$

C3​H4​

C2​H4​

C2​H6​

C3​H6​

7.5mlCx​Hy​​+36ml(X+4Y​)O2​​→XCO2​+2Y​H2​O 36−7.5(x+4y​)+7.5x=28.5 36−7.5(15+4y​)+7.5x=28.5 y=4 x=2 So formula =C2​H4​

$C_{3} H_{4}$

$C_{2} H_{4}$

$C_{2} H_{6}$

$C_{3} H_{6}$

$7.5 m l C_{x} H_{y} + 36 m l (X + \frac{Y}{4}) O_{2} \to XC O_{2} + \frac{Y}{2} H_{2} O$
$36 - 7.5 (x + \frac{y}{4}) + 7.5 x = 28.5$
$36 - 7.5 (15 + \frac{y}{4}) + 7.5 x = 28.5$
$y = 4$
$x = 2$
So formula $= C_{2} H_{4}$