Question

1 mol of a gaseous aliphahatic compound $C_n{H}_{2n}{O}_m$ 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. $(1+\frac{3}{4}n-\frac{1}{2}m)$

Answer 1

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

The general equation for gas analysis of this compound would be

$C_n{H}_{3n}{O}_n\left(g\right)+(n+3/4n-m/2)O_2\left(g\right){⟶}_{n}C{O}_2\left(g\right)+3/2_n{H}_{2}O\left(l\right)$

The contraction in volume of gas can be given as:

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

$=1+\frac{3}{4}n-\frac{m}{2}$.