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How many milligrams of [tex] \text{AlCl}_3 [/tex] (133.5 g/mol) are required to prepare one liter of a solution that is 15 ppm in [tex] \text{Cl} [/tex] (35.5 g/mol)?

A. 18.8 mg
B. 13.8 mg
C. 37.6 mg
D. 27.5 mg

Answer :

Final answer:

To prepare one liter of a solution that is 15 ppm in Cl, 18.8 mg of AlCl3 is required.

Explanation:

First, we calculate the amount of Cl required in milligrams for a 1-liter solution at 15 ppm concentration:

  • 1 liter solution = 1,000,000 milligrams (since 1 liter of water is approximately 1,000 grams and there are 1,000 milligrams in a gram)
  • 15 ppm of Cl means 15 milligrams of Cl in 1,000,000 milligrams of solution

To find out how many milligrams of AlCl3 are needed for 15 milligrams of Cl, we use the molar mass of both AlCl3 and Cl:

  • Molar mass of Cl = 35.5 g/mol
  • Molar mass of AlCl3 = 133.5 g/mol
  • There are 3 moles of Cl in one mole of AlCl3, so 3 * 35.5 g = 106.5 g of Cl in 133.5 g of AlCl3

Now we set up a proportion:

  • (15 mg Cl / 106.5 g Cl) = (x mg AlCl3 / 133.5 g AlCl3)

Solving for x gives us the mass of AlCl3 required:

  • x = (15 mg Cl * 133.5 g AlCl3) / 106.5 g Cl
  • x = 18.8 mg AlCl3

Therefore, the answer is 18.8 mg of AlCl3 required to prepare one liter of a solution that is 15 ppm in Cl.

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Rewritten by : Barada

Final answer:

To prepare a 15 ppm Cl solution in 1 L, approximately 27.5 mg of AICI3 is required.


Explanation:

To calculate the amount of AICI3 required, we can first convert ppm (parts per million) to a weight/volume ratio. Since 1 ppm is equal to 1 mg/L, a 15 ppm solution of Cl would contain 15 mg of Cl in 1 L of solution. Since the molar mass of Cl is 35.5 g/mol, we can set up the following equation to find the amount of AICI3:

AICI3 (g/mol) = (15 mg/L) / (35.5 mg/mol)

Simplifying this equation gives us:

AICI3 = (15/35.5) * (133.5)

Calculating this gives us approximately 56 mg. Therefore, the correct answer is d. 27.5 mg.


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