Question

1 mol of a gaseous aliphatic compound $C_n{H}_{3n}{O}_n$ is completely burnt in an excess of oxygen. The contraction in volume is: (assume water get condensed out)

• 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. $(1+\frac{3}{4}n-\frac{1}{2}n)$.

Answer 1

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

The contraction is due to formation of $C{O}_2$ and $H_{2}O$

$C_n{H}_{3n}{O}_n+n+\frac{3n}{4}-\frac{n}{2}⟶nC{O}_2+\frac{3n}{2}{H}_{2}O$

As the water is condesnded out and thurned out in liquid so we do not calculate it.

Contraction in volume $=1+n+\frac{3n}{4}-\frac{n}{2}-n=1+\frac{3n}{4}-\frac{n}{2}$