Calculate the ionic strength of the following solutions containing 50 mm nacl, 1 mm aic13, 1 mm a12(so4)3.

50 mM NaCl     I = 0.05 M

1 mM AICl₃        I = 6 × 10⁻³ M

1 mM Al₂(SO₄)₃ I = 0.02 M

Explanation:

Ionic strength (I) is a measure of the concentration of ions in solution and can be calculated with the following expression:

I = 1/2 ∑Ci × Zi²

where,

Ci is the molarity of each ion (molarity of the salt × number of each ion in the salt)

Zi is the charge of each ion

50 mM NaCl

NaCl → Na⁺ + Cl⁻

I = 1/2 [C(Na⁺) × Z(Na⁺)² +  C(Cl⁻) × Z(Cl⁻)² ]

I = 1/2 [50 × 10⁻³ M × (+1)² +  50 × 10⁻³ M × (-1)² ]

I = 0.05 M

1 mM AICl₃

AICl₃ → Al⁺³ + 3 Cl⁻

I = 1/2 [C(Al⁺³) × Z(Al⁺³)² +  C(Cl⁻) × Z(Cl⁻)² ]

I = 1/2 [1 × 10⁻³ M × (+3)² +  3 × 10⁻³ M × (-1)² ]

I = 6 × 10⁻³ M

1 mM Al₂(SO₄)₃

Al₂(SO₄)₃ → 2 Al⁺³ + 3 SO₄²⁻

I = 1/2 [C(Al⁺³) × Z(Al⁺³)² +  C(SO₄²⁻) × Z(SO₄²⁻)² ]

I = 1/2 [2 × 10⁻³ M × (+3)² +  3 × 10⁻³ M × (2-)² ]

I = 0.02 M

the volume increases

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.