Question

# One mole of mixture of $$N_2, NO_2$$ and $$N_2O_4$$ has a mean molar mass of $$55.4\ g$$. On the heating to a temperature, at which all the $$N_2O_4$$ may be dissociated into $$NO_2$$, the mean molar mass tends to a lower value of $$39.6\ g$$. What is the molar ratio of $$N_2, NO_2$$ 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

Solution

## 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 $$-y$$$$\dfrac {m_x}{m_x+m_y}=\dfrac {m_y}{m_x+m_y}$$$$\dfrac {m_x}{x}=\dfrac {m_y}{y}$$$$\dfrac {m_x}{m_y}=\dfrac {x}{y}$$mass $$\%$$ of solute $$=\left (\dfrac {m_x}{m_x +m_y}\right)\times 100$$$$=\left (\dfrac {m_x/m_y}{\dfrac {m_x}{m_y}+1}\right)\times 100$$$$=\left (\dfrac {x/y}{\dfrac {x} {y}+1}\right)\times 100$$$$=\left (\dfrac {x/y}{\dfrac {x+y}{y}}\right)\times 100$$$$\therefore \$$ covered $$=\dfrac {x}{x+y}\%$$Let No. of moles of $$N_2=x$$No. of moles of $$NO_2=y$$No. of moles of $$N_2O_4=z$$we get $$x+y+z=1---(1)$$Axerage molar mass $$=55.4$$Molar mass of $$N_2-28, NO_2=46\ N_2O_4=92$$$$\dfrac {28x+46y+92z}{x+y+z}=55.6$$or $$x+y+z=1$$$$\therefore = 28x46y+92z=55.6---(2)$$on heating $$N_2O_4\to 2NO_2$$$$NO_2=2z$$$$\therefore \$$ Total $$NO_2\left (y+2z\right)$$$$\dfrac {28x+46y+92z}{x+y+2z}=39.6$$putting the value from equation $$(2)$$$$x+y+2z=\dfrac {55.4}{39.6}=1.39---(3)$$Now subtract equation $$(1)$$ from equation $$(3)$$$$x+y+2z=1.3y$$$$-x-y-z=-1$$$$z=0.4$$$$\therefore \ x+y=0.6$$putting the value of $$z$$ in equation $$(2)$$$$28x+46y=18.8$$$$-x-y=0.6\times 28$$$$\Rightarrow \ 1.8y=2$$$$\boxed {x=0.1}$$$$\therefore \ x+y=0.6$$$$x+0.1=0.6$$$$\boxed {x=0.5}$$$$x=0.5\quad y=0.1\quad z=0.4$$$$N_2\ :\ NO_3\ :\ N_2O_4$$$$0.5\ :\ 0.1\ :\ 0.4\ \times 10$$$$5:1:4$$$$z=0.4$$Chemistry

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