1.

A hydrocarbon contains 10.5 g of carbon per gram of hydrogen. 1 L of vapour of the hydrocarbon at 127^(@)C and 1 atm pressure weighs 2.8 g. Find the molecular formula of the hydrocarbon.

Answer»

<P>

Solution :Volume at NTP`= (p_(1)V_(1))/(T_(1)) xx (T_(2))/(P_(2))`
`=(1 xx 1)/(400) xx (273)/(1) = (273)/(400)` litre.
`because (273)/(400)` litre weighs 2.8g at NTP
`therefore` 22.4 litres (1 mole) weights `=(2.8 xx 400)/(273) xx 22.4g`
=91.89g `~~92g`
`therefore` molecular weight of the HYDROCARBON =92
`therefore{(,"Weight","Mole",),(C,10.5g,10.5//12,=0.875),(H,1g,1//1,=1):}` or `C:H= 0.875:1`
`=0.875 xx 8: 1 xx`
`=7: 8`
Thus empirical formula is `C_(7)H_(8)`
Now empirical formula weight =92= molecular weight
Hence molecular formula is also `C_(7)H_(8)`


Discussion

No Comment Found

Related InterviewSolutions