Question

10 mL of a gaseous compound, $C_n{H}_{2n+2}{O}_m$ was mixed with 60 mL of $O_2$ and sparked. The gaseous mixture on cooling occupied 45 mL. After treatment with $KOH$ solution the volume of gas remaining was 15 mL. The correct formula is:

Answer 1

$so{l}^n:$ 10 ml of compound $C_n{H}_{2n+2}{O}_m$ is mixed with 60 ml of $O_2$

$C_n{H}_{2n+2}{O}_m+(\frac{7n}{4}-\frac{m}{2})O_2\to nC{O}_2+\left(\frac{2n+2}{2}\right)H_{2}O$

After combination $C_n{H}_{2n+2}{O}_m$ is compound and total volume = 45 ml

Since $C{O}_2$ is acidic, $KOH$ abroes $C{O}_2$ and volume

remaining = 15 ml $=V_{O_2}$ (excerr)

$\therefore V_{C{O}_2}=(45-15)ml=30ml$

Then, after treating with KOH the volume of gas remaining is 15 ml.

$\therefore (60-15)ml=45ml$ of $O_2$ has reacted

Acc. to Avogadro law,

ratio of moles = ratio of volume

$\therefore \frac{O_2}{C{O}_2}=\frac{\frac{zn}{2}-\frac{m}{2}}{n}=\frac{45}{30}$

$45n=30(\frac{7n}{2}-\frac{m}{2})$

$\therefore n=\frac{1}{4}m=0.25m$

$C_{0.25m}{H}_{0.50m+2}{O}_m\equiv C_n{H}_{2n+2}{O}_m$

if $m=1$ then, formula is $C_{0.25}{H}_{2.5}O$