How many milliliters of an aqueous solution of 0.222 M manganese(II) iodide is needed to obtain 9.57 grams of the salt?

139 mL

Explanation:

Given data:

Volume of solution required = ?

Molarity of solution = 0.222 M

Mass of salt = 9.57 g

Solution:

Number of moles of manganese iodide:

Number of moles = mass/molar mass

Number of moles = 9.57 g/ 308.75 g/mol

Number of moles = 0.031 mol

Volume required:

Molarity = number of moles of solute / volume in L

0.222 M = 0.031 mol / V

V = 0.031 mol / 0.222 mol/L

V = 0.139 L

Volume in mL:

0.139 L ×1000 mL / 1 L

139 mL

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