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Calculate the amount of benzoic acid $(C_6H_5COOH)$ required for preparing 250 mL of 0.15 M solution in methanol.

$\begin{array}{1 1}4.575g\\5.575g\\2.575g\\3.575g\end{array} $

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0.15 M solution means that 0.15 moles of $C_6H_3COOH$ is present in 1L=1000mL of the solution
Molar mass of $C_6H_5COOH=72+5+12+32+7=122 gmol^{-1}$
$\therefore$ 0.15 mole of benzoic acid =$0.15\times 122g=18.3g$
Thus,1000mL of solution contains benzoic acid = 18.3g
$\therefore$ 250mL of solution will contain benzoic acid =$\large\frac{18.3}{1000}$$\times 250$
$\Rightarrow 4.575g$
answered Aug 5, 2014 by sreemathi.v

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