Question

1 ml of gaseous aliphatic compound $C_n{H}_{3n}{O}_m$ is completely burnt in an excess of $O_2$ and cooled to room temperature. The contraction in volume is:

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

Answer 1

Answer: (D)

$(1+\frac{3}{4}n-\frac{1}{2}m)$

Solution:- (D) $(1+\frac{3n}{4}-\frac{m}{2})$

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

Volume $=1mL$

As we know that the balanced chemical reaction for the combustion is-

$C_x{H}_y{O}_z+(x+\frac{y}{4}-\frac{z}{2})O_2⟶xC{O}_2+\frac{y}{2}{H}_{2}O$

Here,

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

Therefore, the balanced chemical reaction for the combustion of given organic compound will be-

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

The volume of $H_{2}O$ is ignored as it becomes liquid,

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

Hence the contraction in volume is $(1+\frac{3n}{4}-\frac{m}{2})$.