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10 " mL of " a gaseous organic compound containing C, H and O only was mixed with 100 " mL of " `O_(2)` and exploded under condition which allowed the `H_(2)O` formed to condense. The volume of the gas after explosion was 90 mL. On treatment with KOH solution, a further contraction of 20 mL in volume was observed. The vapour density of the compound is 23. All volume measurements were made under the same condition. Q.The molecular formula of the compound isA. `C_(3)H_(8)O`B. `C_(3)H_(6)O`C. `C_(2)H_(6)O`D. `C_(2)H_(4)O` |
Answer» Correct Answer - C Volume of `O_(2)` reacted with 10 " mL of " compound `=100-70=30mL` Let the formula of the compound be `C_(x)H_(y)O_(z)`. `undersetunderset(10 mol)(10mL)(C_(x)H_(y)O_(z))toundersetunderset(30mL)(30mL)(O_(2))toundersetunderset(20 mol)(20 mL)(xCO_(2))+(y)/(2)H_(2)O` Applying POAC (principle of atom conservation) for C atoms. `x xx` moles of `C_(x)H_(y)O_(z)=1xx` " mol of "`CO_(2)` `x xx10=1xx20` `x=2` Again applying POAC for H and O atoms. `yxx` moles of `C_(x)H_(y)O_(z)=2xx` moles of `H_(2)O` `yxx10=2xx` moles of `H_(2)O` `zxx` moles of `C_(x)H_(y)O_(z)+2xx` moles of `O_(2)` `10z+2xx30=2xx20+` moles of `H_(2)O` ...(ii) Eliminating mols of `H_(2)O` from equation (i) and (ii) we get `y-2z=4` Now, molecular weight of the compound`=2xx23` `=46` `thereforeC_(x)H_(y)O_(z)=2xx12+yxx1+zxx16=46` `y+16z=22` Solve for y and z from equation (iii) and (iv) we get `y=6,z=1` Formula of the compound`=C_(2)H_(6)O` |
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