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The amount of acetic acid (CH₃COOH) in grams needed to prepare the solution is 90

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 CH₃COOH

  • Molarity = 1.5 M
  • Volume = 1 L
  • Mole of CH₃COOH =?

Molarity = mole / Volume

Cross multiply

Mole = Molarity × Volume

Mole of CH₃COOH = 1.5 × 1

Mole of CH₃COOH = 1.5 mole

How to determine the mass of CH₃COOH

  • Mole of CH₃COOH = 1.5 mole
  • Molar mass of CH₃COOH = 60 g/mol
  • Mass of CH₃COOH =?

Mole = mass / molar mass

Cross multiply

Mass = mole × molar mass

Mass of CH₃COOH = 1.5 × 60

Mass of CH₃COOH = 90 g

Thus, 90 g of CH₃COOH is needed to prepare the solution

Learn more about molarity:

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