JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Some Basic Concepts of Chemistry

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}

32 g mol^{- 1}

Answer: 2

Solution

Consider one litre of sea water, whose mass is

1000 \times 2 = 2000 g

. It holds

6 mol

of sodium chloride, that is

6 \times 58.5 = 351 g

, leaving

2000 - 351 = 1649 g

of water as the solvent. A dissolved-oxygen level of

5.8

ppm means

5.8 g

of oxygen per

1 0^{6} g

of solution, so the oxygen present is

\frac{5.8 \times 2000}{1 0 ^{6}} = 1.16 \times 1 0^{- 2} g

, equal to

\frac{1.16 \times 1 0 ^{- 2}}{32} = 3.625 \times 1 0^{- 4} mol

. The molality is

\frac{3.625 \times 1 0 ^{- 4}}{1.649} = 2.2 \times 1 0^{- 4} mol k g^{- 1}

, so to the nearest integer

x

2

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Integer answer