One mole of a mixture of N2-NO2 and N2O4 has a mean molar mass of 55-4- On heating to a temperature at which N2O4 is completely dissociated- N2O4- 2NO2- the mean molar mass tends to a lower value of 39-57- What is the mole ratio of N2-NO2-N2O4 in the original mixture-

Let n_N_2=x n_NO_2=y n_N_2O_4=z Total moles =1 x+y+z=1 (i) mean molar mass before the reaction: 28x+46y+92zx+y+z=55.4 (ii) Once the reaction happens, all ...

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Standard XI

Chemistry

Mole Fraction

One mole of a...

Question

One mole of a 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 $N_{2} O_{4}$ is completely dissociated: $N_{2} O_{4} \to 2 N O_{2}$ , the mean molar mass tends to a lower value of $39.57$ . What is the mole ratio of $N_{2} : N O_{2} : N_{2} O_{4}$ in the original mixture?

0.4 : 0.2 : 0.4

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0.5 : 0.1 : 0.4

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0.6 : 0.2 : 0.2

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0.4 : 0.4 : 0.2

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Solution

The correct option is B 0.5 : 0.1 : 0.4
Let
$n_{N_{2}} = x$
$n_{N O_{2}} = y$
$n_{N_{2} O_{4}} = z$
Total moles $= 1$
$x + y + z = 1 (i)$
mean molar mass before the reaction:
$\frac{28 x + 46 y + 92 z}{x + y + z} = 55.4 (i i)$
Once the reaction happens, all the `z` moles of `N_2O_4` get converted into `2y` moles of `NO_2`
mean molar mass after the reaction:
$\frac{28 x + 46 (y + 2 z)}{x + y + 2 z} = 39.57 (i i i)$
Using `(ii)` and `(iii)`, we get `z=0.4`
Also, we get
$x = 0.5$
$y = 0.1$

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One mole of a mixture of N2,NO2 and N2O4 has a mean molar mass of 55.4. On heating to a temperature at which N2O4 is completely dissociated: N2O4→2NO2, the mean molar mass tends to a lower value of 39.57. What is the mole ratio of N2:NO2:N2O4 in the original mixture?