i got test tomorrow!!!?

Let x mol/L be the solubility of Bi₂S₃, and y mol/L be that of HgS.

Consider the dissolution of Bi₂S₃ :
Bi₂S₃(s) + aq ⇌ 2Bi³⁺(aq) + 3S²⁻(aq)
At equilibrium : [Bi³⁺] = 2x mol/L and [S²⁻] = 3x mol/L
Ksp(Bi₂S₃)= [Bi³⁺]² [S²⁻]³
1 × 10⁻⁹⁷ = (2x)²(3x)³
108x⁵ = 1 × 10⁻⁹⁷
x = 1.6 × 10⁻²⁰
Solubility of Bi₂S₃ = 1.6 × 10⁻²⁰ mol/L

Consider the dissolution of HgS :
HgS(s) + aq ⇌ Hg²⁺(aq) + S²⁻(aq)
At equilibrium : [Hg²⁺] = [S²⁻] = y mol/L
Ksp(HgS) = [Hg²⁺] [S²⁻]
4 × 10⁻⁵³ = y²
y = 6.3 × 10⁻²⁷
Solubility of HgS = 6.3 × 10⁻²⁷ mol/L < 1.6 × 10⁻²⁰ mol/L = Solubility of Bi₂S₃

HgS is less soluble.