Achemist prepares repares a solution of magnesium chloride (mgcl2) by measuring out 48. mg of mgcl, into a 300 ml. volumetric flask and filling to the mark with distilled water.
calculate the molarity of cl^- anions in the chemist's solution.
be sure your answer is rounded to 2 significant digits.

The molarity of the Cl- anions is 0.0033 M

Explanation:

Step 1: Data given

Mass of MgCl2 = 48 mg = 48 *10^-3 grams

volume = 300 mL = 0.3L

Molar mas of MgCl2 = 95.21 g/mol

Step 2: The balanced equation

MgCl2 → Mg2+ + 2Cl-

This means for 1 mol MgCl2, we'll have 2 moles of Cl-

Step 3: Calculate moles MgCl2

Number of moles of MgCl2 = mass of MgCl2 / Molar mass of MgCl2

Moles MgCl2 = 48*10^-3 grams / 95.21 g/mol

Moles MgCl2 = 5.04 *10-4 moles

Step 4: Calculate moles of Cl-

This means for 1 mol MgCl2, we'll have 2 moles of Cl-

For  5.04 *10-4 moles MgCl2, we will have 2* 5.04 *10-4 moles = 0.001 moles Cl-

Step 5: Calculate molarity of the Cl- anion

Molarity = Moles / volume

Molarity Cl- = 0.001 moles / 0.3 L

Molarity Cl- = 0.0033 M

The molarity of the Cl- anions is 0.0033 M

they are both synthetic elements

explanation:

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