A) \[{{C}_{3}}{{H}_{6}}{{O}_{3}}\]
B) \[{{C}_{2}}{{H}_{4}}{{O}_{2}}\]
C) \[{{C}_{2}}{{H}_{4}}O\]
D) \[C{{H}_{2}}O\]
Correct Answer: B
Solution :
Empirical Formula = \[C{{H}_{2}}O\] Empirical formula mass = 12 + 2 + 16 = 30 Mol. Mass = 2 × V.D. = 2 × 30 = 60 \[n=\frac{\text{Mol}\text{.}\,\text{mass}}{\text{Emperical}\,\,\text{mass}}=\frac{60}{30}=2\] Molecular formula = \[{{\text{(Emperical formula)}}_{n}}\] = \[{{(C{{H}_{2}}O)}_{2}}\] = \[{{C}_{2}}{{H}_{4}}{{O}_{2}}\].
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