Hi! May i ask how to solve this qns?
Modified 2010 Alevel P3 Q3 (d)
Alcohol J, CxHyOH, is a volatile fungal metabolite whose
presence when detected in air can indicate hidden fungal attack on
the timbers of a house.
When 0.10cm^3 of liquid J was dissolved in an inert solvent and
an excess of sodium metal added, 10.9cm^3 of hydrogen gas (measured
at 298 k) was produced according to the following equation:
CxHyOH+Na ---> CxHyONa+1/2 H2
When a 0.10cm^3 of liquid J was combusted in an excess of oxygen
in an enclosed vessel, the volume of gas (measured at 298K) was
reduced by 54.4cm^3. The addition of an excess of NaOH caused a
reduction in gas volume of 109cm^3 (measured at 298K)
Use the data to calculate the values of x and y in the molecular
formula for J, CxHyOH.
Balance the equation in terms of x and y. Coefficients should be 2
and (2x + (y-1)/2) --> 2x and (y+1).
If the alcohol was gaseous, based on the redox acid-base reaction
between the monoprotic alcohol and electropositive sodium metal,
determine the volume of gaseous alcohol (ie. if the liquid alcohol
was present in the gaseous state) to be 21.8cm3. This is done out
of convenience, to work in gaseous volumes instead of moles. If you
prefer, you can convert everything to moles instead of working in
gaseous volumes.
Hence, based on CO2 generated, and based on the stoichiometry of
the balanced equation, we have 21.8x = 109, hence x = 5.
Reduction of 54.4 cm3 implies : [Used Up gaseous volume] -
[Generated gaseous volume] = 54.4 cm3.
Alternatively, reduction of 54.4 cm3 implies : -[Used Up gaseous
volume] + [Generated gaseous volume]= -54.4 cm3.
Used up gaseous volume refers to oxygen (since the alcohol was
actually liquid, ie. zero gaseous volume), which is [ (21.8 / 2)(10
+ (y-1)/2) ] cm3, based on the stoichiometry of the balanced
equation.
Generated gaseous volume refers to carbon dioxide, ie. 109
cm3.
Hence y = 11.
