Question

calculate the molarity of ch3cooh in the solution prepared by dissolving its 2.5 g in enough benzene to form 75 ml of the solution?​

Answer 1

Answer :-

Molarity of the solution is 0.554 M .

Explanation :-

We have :-

→ Mass of acetic acid = 2.5 g

→ Volume of solution = 75 mL = 0.075 L

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Firstly, let's calculate molar mass of acetic acid [CH₃COOH] .

= 12 + 1(3) + 12 + 16 + 16 + 1

= 12 + 3 + 45

= 15 + 45

= 60 g/mol

Moles of CH₃COOH :-

= Given Mass/Molar mass

= 2.5/60

= 25/600

= 0.0416 mole

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We have got the required moles of acetic acid (solute) and volume of the solution is given. So now let's calculate the molarity .

Molarity of the solution :-

= Moles of solute/Liters of solution

= 0.0416/0.075

= 416/750

= 0.554 M