Sea water,which can be considered as a 6 M s | vedclass

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(A) Sea water is $6 \ M$ in $NaCl$,so $1000 \ mL$ of sea water contains $6 \ mol$ of $NaCl$.$	ext{Mass of solution} = 	ext{Volume} 	imes 	ext{dens

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$2$

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(A) Sea water is $6 \ M$ in $NaCl$,so $1000 \ mL$ of sea water contains $6 \ mol$ of $NaCl$.$	ext{Mass of solution} = 	ext{Volume} 	imes 	ext{density} = 1000 	imes 2 = 2000 \ g$.$	ext{ppm} = \frac{	ext{mass of } O_2}{	ext{mass of solution}} 	imes 10^6$.$5.8 = \frac{	ext{mass of } O_2}{2000} 	imes 10^6$ $\Rightarrow 	ext{mass of } O_2 = 5.8 	imes 2 	imes 10^{-3} = 1.16 	imes 10^{-2} \ g$.$	ext{Molality of } O_2 = \frac{	ext{moles of } O_2}{	ext{mass of solvent in } kg}$.$	ext{Mass of solvent} = 	ext{Mass of solution} - 	ext{Mass of } NaCl = 2000 - (6 	imes 58.5) = 2000 - 351 = 1649 \ g = 1.649 \ kg$.$	ext{Moles of } O_2 = \frac{1.16 	imes 10^{-2}}{32} = 0.03625 	imes 10^{-2} = 3.625 	imes 10^{-4} \ mol$.$	ext{Molality} = \frac{3.625 	imes 10^{-4}}{1.649} \approx 2.197 	imes 10^{-4} \ m$.Thus,$x \approx 2$.

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