A) \[C{{H}_{2}}O\]
B) \[{{C}_{2}}{{H}_{4}}{{O}_{2}}\]
C) \[{{C}_{3}}{{H}_{6}}{{O}_{3}}\]
D) \[{{C}_{4}}{{H}_{8}}{{O}_{8}}\]
Correct Answer: B
Solution :
The molality is given by \[m=\frac{\Delta {{T}_{f}}}{{{k}_{f}}}=\frac{{{1.16}^{o}}C}{{{1.86}^{o}}C/m}=0.624m\] \[\frac{30.0g}{0.800\text{kg}\,\text{solvent}}=\frac{\text{37}\text{.5}}{\text{kg}\,\text{solvent}}\] Hence, 37.5 g is equivalent to 0.624 mol. \[\frac{37.5g}{0.624\,\text{mol}}=60.1\,\text{g/mol}\] The empirical formula weight is 30 g/eq. formula unit. There must be two units per molecule; the formula is \[{{C}_{2}}{{H}_{4}}{{O}_{2}}.\]
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