Mole ConceptHard
Question
A volume of 1 ml of a gaseous aliphatic compound CnH3nOm is completely burnt in an excess of oxygen. The contraction in volume (in ml) is
Options
A.$1 + \frac{1}{2}n - \frac{3}{4}m$
B.$1 + \frac{3}{4}n - \frac{1}{4}m$
C.$1 - \frac{1}{2}n - \frac{3}{4}m$
D.$1 + \frac{3}{4}n - \frac{1}{2}m$
Solution
$C_{n}H_{3n}O_{m} + \left( \frac{7n}{4} - \frac{m}{2} \right)O_{2} \rightarrow nCO_{2} + \frac{3n}{2}H_{2}O(l)$
$1\text{ mL} \left( \frac{7n}{4} - \frac{m}{2} \right)\text{ mL} n\text{ mL} 0$
Hence, contraction in volume $= \left( 1 + \frac{7n}{4} - \frac{m}{2} \right) - n = \left( 1 + \frac{3n}{4} - \frac{m}{2} \right)\text{ mL}$
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