Question

A 390 ml of aqueous solution prepared by dissolving 0.6 g of aluminum nitrate al(no3)3 in water, the concentration of nitrate ion (no3-) is __________m. the molar mass of al(no3)3 is 213 g/mol

Answer 1

The concentration of the nitrate ion, NO₃¯ in the solution given the data is 0.0216 M

What is molarity?

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

How to determine the mole of Al(NO₃)₃

• Mass of Al(NO₃)₃ = 0.6 g
• Molar mass of solute = 213 g/mol
• Mole of Al(NO₃)₃ =?

Mole = mass / molar mass

Mole of Al(NO₃)₃ = 0.6 / 213

Mole of Al(NO₃)₃ = 0.0028 mole

How to determine the molarity of Al(NO₃)₃

• Mole of Al(NO₃)₃ = 0.0028 mole
• Volume = 390 mL = 390 / 1000 = 0.39 L
• Molarity of Al(NO₃)₃=?

Molarity = mole / Volume

Molarity of Al(NO₃)₃ = 0.0028 / 0.39

Molarity of Al(NO₃)₃ = 0.0072 M

How to determine the molarity of NO₃¯

Dissociation equation

Al(NO₃)₃(aq) <=> Al³⁺(aq) + 3NO₃¯(aq)

From the balanced equation above,

1 mole of Al(NO₃)₃ contains 3 moles of NO₃¯

Therefore,

0.0072 M Al(NO₃)₃ will contain = 0.0072 × 3 = 0.0216 M NO₃¯

Thus, the molarity of NO₃¯ in the solution is 0.0216 M

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