Some amount of dichloromethane (CH_2Cl_2) is | vedclass

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(B) $1$. Calculate the molar mass of $CH_2Cl_2$: $12 + 2(1) + 2(35.5) = 85 \ g \ mol^{-1}$.$2$. Calculate the moles of $CH_2Cl_2$ in $0.671141 \ L$ of

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$148$

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(B) $1$. Calculate the molar mass of $CH_2Cl_2$: $12 + 2(1) + 2(35.5) = 85 \ g \ mol^{-1}$.$2$. Calculate the moles of $CH_2Cl_2$ in $0.671141 \ L$ of solution: $	ext{moles} = 	ext{Molarity} 	imes 	ext{Volume} = 2.6 	imes 10^{-3} \ mol \ L^{-1} 	imes 0.671141 \ L = 1.745 	imes 10^{-3} \ mol$.$3$. Calculate the mass of $CH_2Cl_2$ $(x)$: $x = 	ext{moles} 	imes 	ext{molar mass} = 1.745 	imes 10^{-3} \ mol 	imes 85 \ g \ mol^{-1} = 0.1483 \ g$.$4$. Calculate the mass of the solvent $(CHCl_3)$: $	ext{mass} = 	ext{density} 	imes 	ext{volume} = 1.49 \ g \ cm^{-3} 	imes 671.141 \ cm^3 = 1000 \ g$.$5$. Calculate the concentration in $ppm$: $	ext{ppm} = \frac{	ext{mass of solute}}{	ext{mass of solvent}} 	imes 10^6 = \frac{0.1483 \ g}{1000 \ g} 	imes 10^6 = 148.3 \ ppm \approx 148 \ ppm$.

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