At equilibrium:
PbBr2 <---->
Pb2+
+ 2
Br-
s
2s
Ksp = [Pb2+][Br-]^2
4.6*10^-6=(s)*(2s)^2
4.6*10^-6= 4(s)^3
s = 1.048*10^-2 M
Molar mass of PbBr2,
MM = 1*MM(Pb) + 2*MM(Br)
= 1*207.2 + 2*79.9
= 367 g/mol
Molar mass of PbBr2= 367.0 g/mol
s = 1.048*10^-2 mol/L
To covert it to g/L, multiply it by molar mass
s = 1.048*10^-2 mol/L * 367.0 g/mol
s = 3.845 g/L
Here,
Volume is 2.5 L
So,
Mass = s * volume
= 3.845 g/L * 2.5 L
= 9.6 g
Answer: B
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