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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    Gaseous \[{{N}_{2}}{{O}_{4}}\] dissociates into gaseous \[N{{O}_{2}}\]according to the reaction \[{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)\] At 300 K and 1 atm pressure, the degree of dissociation of \[{{N}_{2}}{{O}_{4}}\] is 0.2. If one mole of \[{{N}_{2}}{{O}_{4}}\] gas is contained in a vessel, then the density of the equilibrium mixture is: JEE Main Online Paper (Held On 10 April 2015)

    A)  3.11 g/L

    B)  6.22 g/L

    C)  4.56 g/L

    D)  1.56 g/L

    Correct Answer: A

    Solution :

                     \[{{N}_{2}}{{O}_{4(g)}}2N{{O}_{2(g)}}\] \[t=0\]                  1              0 t = equilibrium   \[1-\alpha \]       \[2\alpha \] where \[\alpha =\]Degree of disociation. Mol. wt Mixture \[=\frac{\left( 1-\alpha  \right)\times {{M}_{{{N}_{2}}{{O}_{4}}}}+2\alpha +{{M}_{N{{O}_{2}}}}}{\left( 1+\alpha  \right)}=76.66\] Now, As per Ideal gas Equation \[PV=nRT\] \[P{{M}_{mixture}}=dRT\] \[\therefore \] \[d=\frac{P{{M}_{mix}}}{RT}=\frac{1\times 76.66}{0.0821\times 300}=3.11g/L\]


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