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Question
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
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Solution
Given: Volume of solution (V) = 250 mL
Molarity (M) = 0.15 M
Molar mass of benzoic acid (C6H5COOH) = 12 × 6 + 5 × 1 + 12 × 1 + 16 × 2 + 1 × 1
= 72 + 5 + 12 + 32 + 1
= 122 g mol−1
Molarity (M) = `"Number of moles (N) "/"Volume in litres (V)"`
Number of moles = M × V
= 0.15 × 0.250
= 0.0375 mol
Mass = Moles × Molar mass
= 0.0375 × 122
= 4.575 g
Thus, 4.575 g of benzoic acid is required to prepare 250 mL of 0.15 M solution in methanol.
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