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Calculate the amount of benzoic acid (C(...

Calculate the amount of benzoic acid `(C_(6) H_(5) COOH)` required for preparing 250 mL of `0.15 M ` solution in methanol.

Text Solution

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`0.15` M solution means that `0.15 ` moles of `C_(6) H_(5) COOH` is present in 1L
`=1000 mL ` of the solution
Molar mass of `C_(6) H_(5) COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol ^(-1)`
`therefore 0.15` mole of benzoic acid `=0.15 xx 122 g =18.3 g`
Thus, 1000 mL of solution contains benzoic acid `=18.3g`
`therefore 250mL` of solution will contain benzoic acid
`= (18.3)/(1000) xx 250=4.575g`
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