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Standard XII

Chemistry

Relating Translational KE and T

On passing ...

Question

On passing $25$ mL of a gaseous mixture of $N_{2}$ and $N O$ over heated $C u$ , $20$ mL of gas remained. The percentage of $N O$ in the mixture is :

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Solution

Reaction:
$2 N O + C u \to 2 C u O + N_{2}$
$2$ mol $1$ mol
Here, volume reduction is $5$ mL so,
mixture ratio is
$N_{2} N O \to N_{2} N_{2}$
$15$ mL $10$ ml $15$ mL $5$ mL
Therefore, $%$ $N O$ in the mixture $= 10 \times \frac{100}{25} = 40$ mL.

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On passing 25 mL of a gaseous mixture of N2 and NO over heated Cu, 20 mL of gas remained. The percentage of NO in the mixture is :

Reaction:2NO+Cu→2CuO+N2 2 mol 1 molHere, volume reduction is 5 mL so, mixture ratio is N2NO→N2N215 mL 10 ml 15 mL 5 mLTherefore, % NO in the mixture =10×10025=40 mL.

On passing $25$ mL of a gaseous mixture of $N_{2}$ and $N O$ over heated $C u$ , $20$ mL of gas remained. The percentage of $N O$ in the mixture is :

Reaction:
$2 N O + C u \to 2 C u O + N_{2}$
$2$ mol $1$ mol
Here, volume reduction is $5$ mL so,
mixture ratio is
$N_{2} N O \to N_{2} N_{2}$
$15$ mL $10$ ml $15$ mL $5$ mL
Therefore, $%$ $N O$ in the mixture $= 10 \times \frac{100}{25} = 40$ mL.