Question

One mole of a mixture of $N_2$, $N{O}_2$ and $N_2{O}_4$ has a mean molar mass of $55.4$ g/mol. On heating to a temperature at which $N_2{O}_4$ may be dissociated as $N_2{O}_4$ $\to$ $2N{O}_2$, the mean molar mass tends to the lower value of $39.6$ g/mol. What is the mole ratio of $N_2$, $N{O}_2$, $N_2{O}_4$ in the original mixture?

• A. $0.5:0.1:0.4$
• B. $0.1:0.4:0.5$
• C. $10:2:8$
• D. $2:8:10$

Answer 1

Answer: (A), (C)

$0.5:0.1:0.4$
$10:2:8$

Let the original mixture contain $x$ moles of $N{O}_2$ and $y$ moles of $N_2{O}_4$.
The number of moles of $N_2$ will be $1-x-y$.
The molar masses of $N_2$, $N{O}_2$, $N_2{O}_4$ are $28$ g/mol, $46$ g/mol and $92$ g/mol respectively.
The average molar mass of the mixture is $55.4$ g/mol.
$55.4=28(1-x-y)+46x+92y$
$55.4=28-28x-28y+46x+92y$
$27.4=18x+64y$ ......(1)
$y$ moles of $N_2{O}_4$ decompose to form $2y$ moles of $N{O}_2$.
The number of moles of $N_2$, $N{O}_2$, $N_2{O}_4$ will be $(1-x-y),(x+2y)$ and $0$ respectively.
The average molar mass of the new mixture is $39.6$ g/mol.
$39.6(1+y)=28(1-x-y)+46(x+2y)+92\left(0\right)$
$39.6+39.6y=28-28x-28y+46x+92y$
$11.6=18x+24.4y$ .....(2)
On substracting equation (2) from (1), we get
$15.8=39.6y$
$y=0.4$
Substitute value of $y$ in equation (1), we get
$18x=27.4-64\left(0.4\right)=0.1$
Thus, the mole ratio of $N_2$, $N{O}_2$, $N_2{O}_4$ in the original mixture is $0.5:0.1:0.4$.
This is also equal to $10:2:8$.