Answer
$K_{p} = 0.105$
$K_{c} = 2.05\times10^{-3}$
Work Step by Step
Given, $K_{p}$ = 0.105
T = (350+273) K = 623 K
$\Delta n = 1+1-1 = 1$
We know, $K_{p} = K_{c}(RT)^{\Delta n}$
Or $K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{0.105}{\frac{0.08206L.atm}{K.mol}\times623 K} = 2.05\times10^{-3}$
