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(B) Given: Concentration of $CHCl_3 = 15 \ ppm$ (by mass).This means $15 \ g$ of $CHCl_3$ is present in $10^6 \ g$ of the solution.Since the solution

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$1.25 	imes 10^{-4} \ M$

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(B) Given: Concentration of $CHCl_3 = 15 \ ppm$ (by mass).This means $15 \ g$ of $CHCl_3$ is present in $10^6 \ g$ of the solution.Since the solution is dilute,we assume the density of the solution is approximately $1 \ g/mL$,so the volume of $10^6 \ g$ of solution is $10^6 \ mL = 1000 \ L$.Molar mass of $CHCl_3 = 12 + 1 + 3 	imes 35.5 = 119.5 \ g/mol$.Number of moles of $CHCl_3 = \frac{15 \ g}{119.5 \ g/mol} \approx 0.1255 \ mol$.Molarity $(M) = \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.1255 \ mol}{1000 \ L} = 1.255 	imes 10^{-4} \ M$.

$A$ sample of drinking water is found to be severely contaminated with chloroform $(CHCl_3)$,which is a suspected carcinogen. The level of contamination is $15 \ ppm$ (by mass). Calculate the molarity of chloroform in the water sample.

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