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 \[PV=nRT\] \[P{{M}_{mixture}}=dRT\] \[\therefore \] \[d=\frac{P{{M}_{mix}}}{RT}=\frac{1\times 76.66}{0.0821\times 300}=3.11g/L\]You need to login to perform this action.
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