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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
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B
1:1:1
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C
1:4:5
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D
1:5:4
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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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