Q21 / 75 · JEE Main 2025 (4 Apr Shift 2) · Chemistry
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 $\mathrm{x} \times 10^{-4} \mathrm{~m}$. $\mathrm{x}=$ _______. (Nearest integer)
Given: Molar mass of NaCl is $58.5 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_2$ is $32 \mathrm{~g} \mathrm{~mol}^{-1}$
Given: Molar mass of NaCl is $58.5 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_2$ is $32 \mathrm{~g} \mathrm{~mol}^{-1}$
Correct answer: 2
Solution
Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl.
mass of solution $=$ Volume $\times$ density
$=1000 \times 2$
mass of solution $=2000 \mathrm{~g}$
$\begin{aligned}<br />& \text { ppm }=\frac{\text { mass of } \mathrm{O}_2}{2000} \times 10^6 \\<br>& \text { mass of } \mathrm{O}_2=5.8 \times 2 \times 10^{-3} \\<br>& \quad=1.16 \times 10^{-2} \mathrm{~g}<br />\end{aligned}$
molality for $\mathrm{O}_2=\frac{1.16 \times 10^{-2} / 32}{(2000-6 \times 58.5)} \times 1000$
$\begin{aligned} & =\frac{1.16 \times 10}{32 \times 1649} \\ = & 0.000219 \\ = & 2.19 \times 10^{-4}\end{aligned}$
Correct answer $\Rightarrow 2$
mass of solution $=$ Volume $\times$ density
$=1000 \times 2$
mass of solution $=2000 \mathrm{~g}$
$\begin{aligned}<br />& \text { ppm }=\frac{\text { mass of } \mathrm{O}_2}{2000} \times 10^6 \\<br>& \text { mass of } \mathrm{O}_2=5.8 \times 2 \times 10^{-3} \\<br>& \quad=1.16 \times 10^{-2} \mathrm{~g}<br />\end{aligned}$
molality for $\mathrm{O}_2=\frac{1.16 \times 10^{-2} / 32}{(2000-6 \times 58.5)} \times 1000$
$\begin{aligned} & =\frac{1.16 \times 10}{32 \times 1649} \\ = & 0.000219 \\ = & 2.19 \times 10^{-4}\end{aligned}$
Correct answer $\Rightarrow 2$
Hence, the correct answer is 2.