Question

You need to make an aqueous solution of .228 M chromium (III) nitrate for an experiment in lab, using a 500 mL volumetric flask. How much solid chromium (III) nitrate should you add? And how many milliliters of an anqeuous solution of 0.123 M potassium iodide is needed to obtain 9.34 grams of salt?

Answer 1

Generalization of the Principle behind this calculation:

Molarity of a solution = Number of moles of solute present in 1000 mL of the solution

Number of moles of solute present in 1000 mL of the solution

= Mass of the solute in gram present in 1000 mL of the solution / Molar mass of the solute

Therefore

Molarity of a solution = Mass of the solute in gram present in 1000 mL of the solution / Molar mass of the solute

This equation may be used to calculate

i) Molarity of a solution

ii) Mass of the solute in gram present in a given volume of a solution

iii) volume of a solution of known Molarity that will contain a specified quantity of the solute etc.

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Molarity of a solution = Mass of the solute in gram present in 1000 mL of the solution / Molar mass of the solute

Rearranging the above equation

Mass of the solute in gram present in 1000 mL of the solution = Molar mass of the solute x Molarity of the solution

Molar mass of chromium (III) nitrate , Cr(NO₃)₃ = (52.0 + 3x 14.0 + 9 x16.0) g/mol = 238 g/mol

Molarity = 0.228 M

Mass of Cr(NO₃)₃in gram present in 1000 mL of the solution = 238 g/mol x 0.228 M = 54.3 g

Therefore

Mass of Cr(NO₃)₃in gram present in 500 mL of the solution = (54.3 g x 500mL) / 1000mL = 27.15 g = 27.2 g

The number of grams of chromium (III) nitrate solid required to prepare 500 mL of 0.228 M chromium (III) nitrate solution is 27.2 g

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Mass of KI in gram present in 1000 mL of the solution = Molar mass of KI x Molarity of the solution

Molar mass of KI = (39.1+ 127) g/mol = 166.1 g/mol

Molarity of the solution = 0.123 M

Therefore

Mass of KI in gram present in 1000 mL of the solution = 166.1 g/mol x 0.123 M = 20.4 g

20.4 g of KI is present in 1000 mL of the solution

In other words

Volume of KI solution that contains 20.4 g KI = 1000 mL

Therefore

Volume of KI solution that will contain 9.34 g KI = (1000 mL x 9.34 g) /20.4 g = 457.8 ml

The number of milliliters of an anqeuous solution of 0.123 M potassium iodide is needed to obtain 9.34 grams of salt = 457.8 mL

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