मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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

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प्रश्न

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

संख्यात्मक
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उत्तर

Given: Volume of solution (V) = 250 mL

Molarity (M) = 0.15 M

Molar mass of benzoic acid (C6​H5​COOH) = 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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पाठ 1: Solutions - 'NCERT TEXT-BOOK' Exercises [पृष्ठ १२७]

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नूतन Chemistry [English] Class 12 ISC
पाठ 1 Solutions
'NCERT TEXT-BOOK' Exercises | Q 2.30 | पृष्ठ १२७
एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 1 Solutions
Exercises | Q 1.30 | पृष्ठ २९

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