JEE Main 2025 Apr 4 Shift 2, Chemistry Q21: Types Concentration And Solubility of Solutions

JEE Main2025Apr 4, Shift 2Chemistry

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 $^{- 1}$

Your answer

Solution

Take 1 L (1000 mL) of sea water as a basis.

Mass of solution $= 1000 \times 2 = 2000$ g, and it contains 6 mol of NaCl, i.e. $6 \times 58.5 = 351$ g of NaCl. Mass of water (solvent) $= 2000 - 351 = 1649$ g.

From the ppm definition (mass/mass $\times 1 0^{6}$ ):

$mass of O_{2} = \frac{5.8 \times 2000}{1 0 ^{6}} = 1.16 \times 1 0^{- 2}$ g.

Moles of $O_{2} = \frac{1.16 \times 1 0 ^{- 2}}{32}$ .

Molality $= \frac{moles of O _{2}}{mass of solvent in kg} = \frac{1.16 \times 1 0 ^{- 2} /32}{1.649} \approx 2.19 \times 1 0^{- 4} m$ .

$∴ x \approx 2$ .

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JEE Main 2025 Apr 4 Shift 2, Chemistry Q21: Types Concentration And Solubility of Solutions

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