Sea water, which can be considered as a 6 molar (6 M) solution of NaCl, has a density of 2 g . The concentration of dissolved oxygen () in sea water is 5.8 ppm. Then the concentration of dissolved oxygen () in sea water, is m. x = _______. (Nearest int

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Solution

Q.
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 $x \times 10^{- 4}$ m.

x = _______. (Nearest integer)

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

Molar mass of $O_{2}$ is 32 g ${mol}^{- 1}$ [2025]

Ans.
(2)

$(i) 6M NaCl solution means 6 mol NaCl in 1000 mL solution$

$i.e. 6 mol NaCl in 1000 mL \times 2 {g mL}^{- 1} solution$

$i.e. 6 mol NaCl in 2000 g solution$

$i.e. 6 \times 58.5 g NaCl in 2000 g solution$

$i.e. 351 g NaCl in 2000 g water.$

${(ii) ppm of O}_{2} = \frac{W_{O_{2}}}{W_{solvent}} \times 10^{6}$

$5.8 = \frac{W_{O_{2}}}{2000} \times 10^{6}$

$W_{O_{2}} = \frac{5.8 \times 2000}{10^{6}} = 0.0116 g$

$(iii) Mass of solvent in 2000 g solution (W_{solvent})$

$= W_{solution} - W_{NaCl} - W_{O_{2}}$

$= 2000 - 351 - 0.0116 = 1648.9884 g$

$Molality of oxygen (m) = \frac{W_{O_{2}} \times 1000}{M_{O_{2}} \times W_{solvent}}$

$= \frac{0.0116 \times 1000}{32 \times 1648.9884} = 2.19 \times 10^{- 4}$

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