Sea water, which can be considered as a 6 molar ( 6 M ) solution of NaCl , has a density of 2 g mL − 1 . The concentration of dissolved oxygen ( O 2 ) in sea water is 5.8 ppm. Then the concentration

Question

Sea water, which can be considered as a 6 molar $(6 M)$ solution of NaCl , has a density of $2 g mL^{- 1}$ . The concentration of dissolved oxygen $(O_{2})$ in sea water is 5.8 ppm. Then the concentration of dissolved oxygen $(O_{2})$ in sea water, is $x \times 1 0^{- 4} m$ . $x =$ _______. (Nearest integer)
Given: Molar mass of NaCl is $58.5 g mol^{- 1}$ Molar mass of $O_{2}$ is $32 g mol^{- 1}$

TestHub · Accepted Answer

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 $= 2000 g$
$ppm = \frac{mass of O _{2}}{2000} \times 1 0^{6} mass of O_{2} = 5.8 \times 2 \times 1 0^{- 3} = 1.16 \times 1 0^{- 2} g$
molality for $O_{2} = \frac{1.16 \times 1 0 ^{- 2} /32}{( 2000 - 6 \times 58.5 )} \times 1000$
$= = = \frac{1.16 \times 10}{32 \times 1649} 0.000219 2.19 \times 1 0^{- 4}$
Correct answer $\Rightarrow 2$

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