1.

A `10 L` box contains `41.4 g` of a mixture of gases `C_(x)H_(8)` and `C_(x)H_(12)`. The total pressure at `44^(@)C` in flask is `1.56 atm`. Analysis revelated that the gas mixture has `87%` total `C` and `13%` total `H`. Find out the value of `x`A. `1`B. `3`C. `5`D. `2`

Answer» Given `P=1.56 atm`, `V=10 L`
`T=317 K`, `R=0.082`
Total moles `(m)=(PV)/(RT)=(1.56xx10)/(0.082xx317)=0.6 mol`
Let `CxH_(8)` be a mol, therefore moles of
`C_(x)H_(12)=(0.6-a)mol`, mas of `C` in a mol of
`C_(x)H_(12)-12ax g`, mass of `C` in `(0.6-a) mol` of
`C_(x)H_(12)=12xx(0.6-a)g`
`:. Total mass of C in mixture=12ax+12x(0.6-a)g`
`=41.4 g`
`%` of `C` in mixture `=(7.2x)/(41.4)xx100`
Given `%` of `C=87%`
or `(720 x)/(41.4)=87` or `x=5`


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