Q 2.30 Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

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Vishal Baghel | Contributor-Level 10

8 months ago

Volume of the solution = 250mL = 0.25L

Let the no. of moles of solute be n

Molarity = No. of moles of solute/volume of solution

⇒ 0.15 = n/0.25

⇒ n = 0.0375moles

Molar mass of C₆H₅OH = 6*12 + 5*1 + 16 + 1 = 94g

Moles = mass/molar mass

⇒ 0.0375 = m/94

Mass of benzoic acid required = 3.525g.

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