A) \[{{C}_{6}}{{H}_{6}}\]
B) \[{{C}_{12}}{{H}_{6}}\]
C) \[{{C}_{7}}{{H}_{8}}\]
D) \[{{C}_{14}}{{H}_{10}}\]
Correct Answer: D
Solution :
| [d] According to Raoult's law |
| \[\frac{\Delta p}{{{p}^{0}}}={{x}_{2}}\] |
| Where \[-\Delta p\] = (74.01 - 74.66) torr and \[{{p}^{0}}\]= 74.66 torr |
| If M is the molar mass of hydrocarbon, then |
| \[{{X}_{2}}=\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=\frac{\frac{8}{M}}{\left( \frac{100}{78} \right)+\left( \frac{2}{M} \right)}\] |
| Hence \[\frac{74.66-74.01}{74.66}=\frac{\frac{2}{M}}{\frac{100}{78}+\frac{2}{M}}\] |
| Solving for M, we get, M = 177.6 g \[mo{{l}^{-1}}\]. |
| Given mass ratio is \[{{m}_{C}}:{{m}_{H}}\] : : 94.4 : 5.6 |
| This atomic ratio is |
| \[{{N}_{C}}:{{N}_{H}}::\frac{84.4}{12} :\frac{5.6}{1}\Rightarrow 7.87 :5.6\] |
| \[\Rightarrow 1.4 :1\Rightarrow 7:5\] |
| Hence, Empirical formula is \[{{C}_{7}}{{H}_{5}}\] |
| Molar Empirical mass = 89 g \[mo{{l}^{-1}}\] |
| Number of \[{{C}_{7}}{{H}_{5}}\] unit in the given molecule \[=\frac{\text{Molar}\,\text{mass}}{\text{Molar}\,\text{empirical}\,\text{mass}}=\frac{177.6}{89}\cong 2\] |
| Thus molecular formula is \[{{C}_{14}}{{H}_{10}}\] |
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