Question

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

• A. $5:1:4$
• B. $1:1:1$
• C. $1:4:5$
• D. $1:5:4$

Answer 1

The correct option is A $5:1:4$

$x_{solute}=x_{solvent}$
$x_{solute}+x_{solvent}=1$
$\therefore x_{solute}=x_{solvent}=0.5$
solute $-x$, solvent $-y$
$\frac{m_x}{m_x+m_y}=\frac{m_y}{m_x+m_y}$
$\frac{m_x}{x}=\frac{m_y}{y}$
$\frac{m_x}{m_y}=\frac{x}{y}$
mass $\%$ of solute $=\left(\frac{m_x}{m_x+m_y}\right)\times 100$
$=\left(\frac{m_x/m_y}{\frac{m_x}{m_y}+1}\right)\times 100$
$=\left(\frac{x/y}{\frac{x}{y}+1}\right)\times 100$
$=\left(\frac{x/y}{\frac{x+y}{y}}\right)\times 100$
$\therefore$ covered $=\frac{x}{x+y}\%$
Let No. of moles of $N_2=x$
No. of moles of $N{O}_2=y$
No. of moles of $N_2{O}_4=z$
we get $x+y+z=1---\left(1\right)$
Axerage molar mass $=55.4$
Molar mass of $N_2-28,N{O}_2=46N_2{O}_4=92$
$\frac{28x+46y+92z}{x+y+z}=55.6$
or $x+y+z=1$
$\therefore =28x46y+92z=55.6---\left(2\right)$
on heating $N_2{O}_4\to 2N{O}_2$
$N{O}_2=2z$
$\therefore$ Total $N{O}_2(y+2z)$
$\frac{28x+46y+92z}{x+y+2z}=39.6$
putting the value from equation $\left(2\right)$
$x+y+2z=\frac{55.4}{39.6}=1.39---\left(3\right)$
Now subtract equation $\left(1\right)$ from equation $\left(3\right)$
$x+y+2z=1.3y$
$-x-y-z=-1$
$z=0.4$
$\therefore x+y=0.6$
putting the value of $z$ in equation $\left(2\right)$
$28x+46y=18.8$
$-x-y=0.6\times 28$
$\Rightarrow 1.8y=2$
$x=0.1$
$\therefore x+y=0.6$
$x+0.1=0.6$
$x=0.5$
$x=0.5y=0.1z=0.4$
$N_2:N{O}_3:N_2{O}_4$
$0.5:0.1:0.4\times 10$
$5:1:4$
$z=0.4$