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प्रश्न
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
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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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\phantom{.......}\ce{CH3}\\
\phantom{....}|\\
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