A) \[{{C}_{5}}{{H}_{12}}\]
B) \[{{C}_{6}}{{H}_{10}}\]
C) \[{{C}_{3}}{{H}_{8}}\]
D) \[{{C}_{4}}{{H}_{10}}\]
Correct Answer: D
Solution :
| [d] Idea This problem can be solved by using concept of ideal gas equation and chemical equation involved in combustion reaction. Students are advised to follow steps. |
| Write chemical equation involved in combustion of hydrocarbon. |
| Assume initial pressure = p |
| Calculate increased pressure |
| Then, calculate number of mole using ideal gas equation. |
| Equation of combustion |
| \[{{C}_{n}}{{H}_{2n}}_{+2}+\left( \frac{3n+1}{2} \right){{O}_{2}}\xrightarrow{{}}nC{{O}_{2}}+(n+1){{H}_{2}}O\] |
| Initial pressure of \[{{C}_{n}}{{H}_{2n}}_{+2}\] is p (assumed) |
| Increase in pressure |
| \[=p\left[ (2n+1)-1-\left( \frac{3n+1}{2} \right) \right]=\left( \frac{n-1}{2} \right)p\] |
| \[\because \] Mass of organic compound \[=14\times n+2\] |
| 546 K and 4.6 atm or 273 K 2.3 atm |
| Increase \[\Rightarrow \] \[2.3-2=0.3\] atm |
| \[p=\frac{nRT}{V}=\frac{23.2}{M}\times \left( \frac{0.0821\times 273}{44.82} \right)\] |
| \[=\frac{23.2}{M}\times \frac{0.5}{44.82}\] |
| \[=\frac{116}{(14n+2)}\times \frac{(n-1)}{2}\] |
| also \[=\frac{(n-1)}{2}\times \frac{11.6}{14n+2}=0.3\] |
| On solving, \[n=4\] |
| So, compound will be \[{{C}_{4}}{{H}_{4\times 2+2}}\Rightarrow {{C}_{2}}{{H}_{10}}\] |
| TEST Edge. JEE Main includes this type of question in exam to judge the knowledge as well as numerical aptitude of students. So, students are advised to solve the question as much as possible. |
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