Question

1 mole of a gaseous burnt in is aliphatic compound $C_n{H}_{3n}{O}_m$ is completely burnt in excess of oxygen. The contraction is in volume is:

• A. $(1+\frac{1}{2}n-\frac{3}{4}m)$
• B. $(1+\frac{3}{4}n-\frac{1}{4}m)$
• C. $(1-\frac{1}{2}n-\frac{3}{4}m)$
• D. $(1+\frac{3}{4}n-\frac{1}{2}m)$

Answer 1

The correct option is D $(1+\frac{3}{4}n-\frac{1}{2}m)$

The gaseous aliphatic organic compound = $C_n{H}_{3n}{O}_m$

Volume= 1 mL

The balanced reaction for the combustion is,

$C_x{H}_y{O}_z+(x+y/4-z/2)O_2\to xC{O}_2+y/2H_{2}O$

Here, x=n, y=3n, z=m

Hence, the balanced reaction for the combustion of given organic compound will be,

$C_n{H}_{3n}{O}_m+(n+3n/4-m/2)O_2\to nC{O}_2+3n/2H_{2}O$

At RT, the volume of H2O is ognored as it becomes liquid,

Hence, contraction in volume = 1 + n + 3n/4 - m/2 - n = $(1+\frac{3}{4}n-\frac{1}{2}m)$