A) 43 g
B) 86 g
C) 129 g
D) 172 g
Correct Answer: A
Solution :
[a] |
Moles of \[C={}^{n}C{{O}_{2}}=\frac{2.64}{44}=0.06\] \[\Rightarrow \] mass of C= 0.72 |
Mole of \[H=2\times Moles\]of \[{{H}_{2}}O=\]\[2\times \frac{1.26}{18}=0.14\Rightarrow \]mass of \[H=0.14\] |
Compound does not contains oxygen. |
So \[EF\to {{C}_{0.06}}{{H}_{0.14}}\Rightarrow {{C}_{3}}{{H}_{17}}\] \[\Rightarrow \] Lowest M.M. = 43 |
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