JEE Main 17 March 2021 Shift 2 Solved Paper

The temperature at which KC=20.4 and Kρ=600.1, is 354K.
Given reaction is, N2O4(g)⇌2NO2(g)
Given values are : Kρ=600.1
KC=20.4
∆ng= number of moles of product - number of moles of reactant Using relation between Kp and KC=2−1=1
where R is the gas constant =0.083Latm/Kmol
∆ng=1 (for given reaction)
On putting given values, we will get
600.1=20.4(RT)1⇒T≈354K