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

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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

V = 250 mL, m = 0.15 M, Molecular weight of solute = 122, Amount of solute = ?

m = `"Weight of solute"/"Molecular weight of solute" xx 1000/"Volume of solution in mL"`

or 0.15 = `w/122 xx 1000/250`

w = `(0.15 xx 122 xx 250)/1000`

= 4.575 g

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अध्याय 1: Solutions - Exercises [पृष्ठ २९]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
अध्याय 1 Solutions
Exercises | Q 1.30 | पृष्ठ २९

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