Question

$0.2$ moles of a hydrocarbon, which cannot decolourise bromine water, on complete combustion produced $26.4$ g of $C{O}_2$. The molecular weight of the hydrocarbon is:

• A. $44$
• B. $42$
• C. $40$
• D. $58$

Answer 1

Answer: (A)

$44$

Since, the hydrocarbon cannot decolorize Br water, it cannot be alkene or alkyne. Therefore, it is an alkane.
Let the molecular formula be $C_n{H}_{2n+2}$
$26.4$ g of $C{O}_2=0.6$ moles
Therefore, $0.2$ moles of hydrocarbon contains $0.6$ moles of carbon.
$1$ mole will contain $\frac{0.6}{0.3}=3$ moles.
So, the formula is $C_3{H}_8$ and molecular weight $=44$ g.