The correct option is B 2.7 M
To dissolve 0.1 mole of AgCl in 1 L, let conc of NH3 be C and when simultaneous equilibria will be established, conc of Cl− will be 0.1 M and of complex will be also almost 0.1 M as Kf is large. Let conc Ag+ left over be x.
AgCl(s)Ksp=10−10⇌Ag+aq+Cl−aq x 0.1
Ag+aq+2NH3Kf=1.6×107⇌[Ag(NH3)2]+
x C−0.1×2 0.1
So, Ksp=10−10=x×0.1⇒x=10−9 M
And Kf=1.6×107=0.1x×(C−0.1×2)2=0.110−9×(C−0.1×2)2
⇒(C−0.1×2)2=10.16⇒C−0.1×2=10.4⇒C=2.7 M