A) 1.56 g/L
B) 6.22 g/L
C) 3.11g/L
D) 4.56 g/L
Correct Answer: C
Solution :
[c] \begin{align} & \begin{matrix} {} & {} & {{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g) \\ \end{matrix} \\ & \begin{matrix} t=0 & {} & 1 \\ \end{matrix}\begin{matrix} {} & {} & {} & 0 \\ \end{matrix} \\ & \begin{matrix} t=\text{eq}\text{.} & 1-\alpha & {} \\ \end{matrix}\begin{matrix} {} & 2\alpha \\ \end{matrix} \\ \end{align}\] Where \[\alpha \] = Degree of dissociation. Mol. wt. of mixture \[=\frac{(1-\alpha )\times {{M}_{{{N}_{2}}{{O}_{4}}}}+2\alpha \times {{M}_{N{{O}_{2}}}}}{(1+\alpha )}\] \[=\frac{(1-0.2)92+2\times 0.2\times 46}{(1+0.2)}=76.66\] Now, As per ideal gas equation \[PV=nRT\] \[P{{M}_{mixture}}=dRT\] \[\therefore d=\frac{P{{M}_{mix}}}{RT}=\frac{1\times 76.66}{0.0821\times 300}=3.11g/L\]You need to login to perform this action.
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