When 0-44 g of a colourless oxide of nitrogen occupies 224 mL at 1520 mm of Hg and 273- - Cl - then the oxide isA- N 2 O 5B- N 2 O 3C- NO 2D- N 2 O

The correct option is D N2O Volume of oxide is calculated at STP conditions by the formula P_1V_1/T_1=P_2V_2/T_2 1520 × 224/546 = 760 × V_2/273 V_2 = 1520 × 22 ...

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Standard X

Chemistry

Mole Concept

When 0.44 g o...

Question

When $0.44 g$ of a colourless oxide of nitrogen occupies $224 m L$ at $1520 m m$ of $H g$ and $273^{\circ}C$ , then the oxide is

N₂O₅

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N₂O₃

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NO₂

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N₂O

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Solution

The correct option is D
N₂O

Volume of oxide is calculated at STP conditions by the formula $\frac{P_{1} V_{1}}{T_{1}}$ = $\frac{P_{2} V_{2}}{T_{2}}$

$\frac{1520 \times 224}{546}$ = $\frac{760 \times V_{2}}{273}$

$V_{2}$ = $\frac{1520 \times 224 \times 273}{546 \times 760}$ =224ml

Weight of the $22400 m l$ oxide at STP = molecular weight

Weight of the $224 m l$ = $0.44 g$

Weight of the $22400 m l$ = $44 g$

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When 0.44 g of a colourless oxide of nitrogen occupies 224 mL at 1520 mm of Hg and \(273^{\circ}C\) , then the oxide is

The correct option is D N2O Volume of oxide is calculated at STP conditions by the formula P1V1T1=P2V2T2 1520 × 224546 = 760 × V2273 V2 = 1520 × 224 × 273546 × 760 =224ml Weight of the 22400ml oxide at STP = molecular weight Weight of the 224ml = 0.44 g Weight of the 22400ml = 44g

When $0.44 g$ of a colourless oxide of nitrogen occupies $224 m L$ at $1520 m m$ of $H g$ and \(273^{\circ}C\) , then the oxide is