Question

How many grams of magnesium chloride must be added to 725 mL of water to create a solution with an anion concentration equal to 0.800 M

Answer 1

55.2218 grams of magnesium chloride must be added to 725 mL of water to create a solution with an anion concentration equal to 0.800 M.

How to calculate the no. of moles ?

To calculate the number of moles from molarity as

$\frac{\text{x moles}\ MgCl_2}{0.725\ \text{L solution}} = \frac{0.8\ \text{moles}\ MgCl_2}{1.0\ \text{L solution}}$

x moles MgCl₂ = $\frac{0.8\ \text{moles}\ MgCl_2}{1.0\ \text{L solution}} \times 0.725\ \text{L solution}$

                         = 0.58 moles MgCl₂

What is Molar Mass ?

Mass of one mole of substance is known as molar mass.

Molar mass of MgCl₂ = Atomic mass of Mg + 2 (Atomic mass of Cl)

                                   = 24.31 + 2 (35.5)

                                   = 95.21

Convert moles to grams

0.58 moles MgCl₂ × $\frac{95.21\ g}{1\ \text{mole}\ MgCl_2}$

= 55.2218 grams

Thus from the above conclusion we can say that 55.2218 grams of magnesium chloride must be added to 725 mL of water to create a solution with an anion concentration equal to 0.800 M.

Learn more about the Molar mass here: brainly.com/question/837939

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