$20 mL$ of a gaseous hydrocarbon $C X_{x} H X_{y}$ is mixed with $100 mL$ of $O X_{2}$ and exploded. After cooling to the initial temperature and pressure (water liquid), the gas volume is $70 mL$ ; after passing through KOH solution it falls to $10 mL$ . Find the molar mass of the hydrocarbon in $g mo l^{- 1}$ (C = 12, H = 1), to the nearest integer.

Numerical answer type, answer in g/mol, tolerance 0.1.

Answer: 42 g/mol

Volume absorbed by KOH

= 70 - 10 = 60

mL of

CO X_{2}

, so

20 x = 60

and

x = 3

. The residual

10

mL is unused

O X_{2}

, so

O X_{2}

consumed

= 90

mL. From

C X_{x} H X_{y} + (x + \frac{y}{4}) O X_{2} x CO X_{2} + (\frac{y}{2}) H X_{2} O

20 (x + y /4) = 90

, so

3 + y /4 = 4.5

and

y = 6

. The hydrocarbon is

C X_{3} H X_{6}

with

M = 36 + 6 = 42 g mo l^{- 1}