10.30 ; mg of O_2 is dissolved in 1 ; L of se | vedclass

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Question

(B) The mass of $O_2$ is $10.30 \; mg = 10.30 	imes 10^{-3} \; g$. The volume of seawater is $1 \; L = 1000 \; mL$. The density of seawater is $1.03

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The mass of $O_2$ is $10.30 \; mg = 10.30 	imes 10^{-3} \; g$. The volume of seawater is $1 \; L = 1000 \; mL$. The density of seawater is $1.03 \; g/mL$. The mass of seawater = $	ext{density} 	imes 	ext{volume} = 1.03 \; g/mL 	imes 1000 \; mL = 1030 \; g$. The concentration in $ppm$ is calculated as: $ppm = \frac{	ext{mass of solute}}{	ext{mass of solution}} 	imes 10^{6}$ $ppm = \frac{10.30 	imes 10^{-3} \; g}{1030 \; g} 	imes 10^{6} = \frac{10.30}{1030} 	imes 10^{3} = 0.01 	imes 1000 = 10$.

$10.30 \; mg$ of $O_2$ is dissolved in $1 \; L$ of seawater with a density of $1.03 \; g/mL$. The concentration of $O_2$ in $ppm$ is:

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