Calculate the volume in liters of of a 7.62x10^-6 umol/L Mercury (II) iodide solution that contains 700 mg of mercury (II) iodide (HgI2). Round answer 3 significant figures.

2.02 * 10⁸ L

Explanation:

Moles is denoted by given mass divided by the molecular mass,

Hence,

n = w / m

n = moles,

w = given mass,

m = molecular mass.

From the question,

w = 700 mg,

(since, 1 mg = 10 ⁻³ g)

hence, w = 0.700 g

m = molecular mass of HgI₂ = 454.4 g/mol.

Calculating moles by using the above formula,

n = w / m = 0.700 g / 454.4 g/mol = 1.54 * 10⁻³ mol.

From the question,

7.62 x 10⁻⁶ umol/L = 7.62 * 10⁻¹² mol / L

Since, 1 umol = 10 ⁻⁶ mol

The Volume in terms of liters is -

volume = 1 L * 1.54 * 10⁻³ mol / 7.62 * 10⁻¹² mol / L = 2.02 * 10⁸ L