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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.

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Answer:

2.02 * 10⁸ L

Step-by-step 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

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