Sea water, which can be considered as a 6 molar $(6 \mathrm{M})$ solution of NaCl , has a density of $2 \mathrm{~g} \mathrm{~mL}^{-1}$. The concentration of dissolved oxygen $\left(\mathrm{O}_2\right)$ in sea water is 5.8 ppm . Then the concentration of dissolved oxygen $\left(\mathrm{O}_2\right)$ in sea water, is $x \times 10^{-4} \mathrm{~m}$.

$x=$ ___________. (Nearest integer)

Given: Molar mass of NaCl is $58.5 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of $\mathrm{O}_2$ is $32 \mathrm{~g} \mathrm{~mol}^{-1}$

<p>Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl .</p> <p>$$\begin{aligned} & \text { mass of solution }=\text { Volume } \times \text { density } \\ & =1000 \times 2 \\ & \text { mass of solution }=2000 \mathrm{~g} \\ & \mathrm{ppm}=\frac{\text { mass of } \mathrm{O}_2}{2000} \times 10^6 \\ & \text { mass of } \mathrm{O}_2=5.8 \times 2 \times 10^{-3} \\ & \quad=1.16 \times 10^{-2} \mathrm{~g} \\ & \text { molality for } \mathrm{O}_2=\frac{1.16 \times 10^{-2} / 32}{(2000-6 \times 58.5)} \times 1000 \end{aligned}$$</p> <p>$$\begin{aligned} &\begin{aligned} & =\frac{1.16 \times 10}{32 \times 1649} \\ & =0.000219 \\ & =2.19 \times 10^{-4} \end{aligned}\\ &\text { Correct answer } \Rightarrow 2 \end{aligned}$$</p>