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A 10 L box contains 41.4 g of a mixture ...

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 1.56 atm. Analysis revealed that the gas mixture has 87% total C and 13% total H. Find out the value of x

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The correct Answer is:
5
Given , `P = 1.56` atm , `V = 10 L , T = 317 K , R = 0.082`
Total moles (n) = `(PV)/(RT) = (1.56 xx 10)/(0.082 xx 317) = 0.6` mol
Let `C_(x) H_8` be .a. mole , therefore moles of `C_(x) H_(12) = (0.6 - a)` mol , mass of C in .a. mol of `C_(x) H_(8) = 12` ax g , mass of C in (0.6 - a) mol of `C_(x) H_(12) = 12 x (0.6 - a) g`
`therefore` Total mass of C in mixture = 12 ax + 12 ax (0.6 - a) g
% of C in mixture = `(7.2 x)/(41.4) xx 100` , Given % of C= 87 % or `(720 x)/(41.4) = 87` or x = 5
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