1 mol mixture of N2- NO2 and N2O4 has a mean molar mass of 55-4- On heating to a temperature at which all the N2O4 dissociated as - N2O4 - 2NO2 The mean molar mass of mixture becomes 39-6- The molar ratio of NO2 in the original mixture is -

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Mole Concept

1 mol mixture...

Question

$1$ mol mixture of $N_{2}, N O_{2}$ and $N_{2} O_{4}$ has a mean molar mass of $55.4$ . On heating to a temperature at which all the $N_{2} O_{4}$ dissociated as :

$N_{2} O_{4} \to 2 N O_{2}$

The mean molar mass of mixture becomes $39.6$ . The molar ratio of $N O_{2}$ in the original mixture is :

0.4

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0.3

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0.1

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Solution

The correct option is D 0.1
Let $a, b$ and $c$ $m o l$ of $N_{2}, N O_{2},$ and $N_{2} O_{4}$ are present in $1$ $m o l$ of a mixture.

Thus,
$a + b + c = 1$ $. . . . (i)$
Also,
$\frac{(a \times 28) + (b \times 46) + (c \times 92)}{a + b + c} = 55.4$
$∴ 28 a + 46 b + 92 c = 55.4$ $. . . . (i i)$
Also, after heating,
$\frac{(a \times 28) + [(b + 2 c) \times 46]}{a + b + 2 c} = 39.6$ $. . . . (i i i)$
$∴ 28 a + 46 b + 92 c = (a + b + 2 c) \times 39.6$ $. . . . (i v)$
From equations $(i i)$ and $(i v)$ , we get
$a + b + 2 c = \frac{55.4}{39.6} = 1.40$ $. . . . (v)$
From equations $(i i)$ and $(v)$ , we get,

$28 a + 46 b + 92 \times 0.4 = 55.4$

$28 a + 46 b = 18.6$ $. . . . (v i)$

From equation $(i)$ and $(v i)$ $a + b = 0.6, b = 0.1, a = 0.5$

$∴ N_{2} = 0.5, N O_{2} = 0.1, N_{2} O_{4} = 0.4$ mol.

Therefore, the option is D.

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1 mol mixture of N2, NO2 and N2O4 has a mean molar mass of 55.4. On heating to a temperature at which all the N2O4 dissociated as : N2O4→2NO2 The mean molar mass of mixture becomes 39.6. The molar ratio of NO2 in the original mixture is :

$1$ mol mixture of $N_{2}, N O_{2}$ and $N_{2} O_{4}$ has a mean molar mass of $55.4$ . On heating to a temperature at which all the $N_{2} O_{4}$ dissociated as :

$N_{2} O_{4} \to 2 N O_{2}$

The mean molar mass of mixture becomes $39.6$ . The molar ratio of $N O_{2}$ in the original mixture is :