Question
A volume of 10 ml of an oxide of nitrogen was taken in a eudiometer tube and mixed with hydrogen until the volume was 28 ml. On sparking, the resulting mixture occupied 18 ml. To this mixture, oxygen was added when the volume came to 27 ml and on explosion again, the volume fall to 15 ml. Find the molecular weight of the oxide of nitrogen originally taken in eudiometer tube. All measurements were made at STP.
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Solution
$N_{x}O_{y} + yH_{2} \rightarrow \frac{x}{2}N_{2} + yH_{2}O$
10 mL 10y mL 5x mL 0
Contraction in volume, (10 + 10y) − 5x = 28 − 18 ⇒ x = 2y
Volume of H2 unreacted = (28 − 10) − 10y
= (18 − 10y) mL
2H2 + O2 → 2H2O
(18 − 10y) mL (9 − 5y) mL 0
Contraction in volume, 27 − 15y = 27 − 15
⇒ y = 1 and x = 2 ⇒ N2O ⇒ Molecular mass = 44
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