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NVAMediumJEE 2025Concentration Terms (Molarity, Molality, etc.)

JEE Chemistry 2025 Question with Solution

Sea water, which can be considered as a 6 M6 \, \text{M} solution of NaCl, has a density of 2 g mL−12 \, \text{g mL}^{-1}. The concentration of dissolved oxygen (O_2$$**) in sea water is **$$5.8 \, \text{ppm}$$**. Then the concentration of dissolved oxygen (**O_2∗∗)inseawater,in∗∗**) in sea water, in **x \times 10^{-6} , \text{m}∗∗.∗∗**. **x$$ = _____. (Nearest integer)

Given: Molar mass of NaCl is 58.5 g mol−158.5 \, \text{g mol}^{-1} Molar mass of O$$$_2∗∗is∗∗** is **32 , \text{g mol}^{-1}$$.

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: Sea water is 6 M6 \, \text{M} in NaCl, density is 2 g mL−12 \, \text{g mL}^{-1}, and dissolved oxygen is 5.8 ppm5.8 \, \text{ppm}.

Find: The value of xx if concentration of dissolved O$$$_2∗∗iswrittenas∗∗** is written as **x \times 10^{-6} , \text{m}$$.

For 1000 mL1000 \, \text{mL} of sea water, mass of solution is

Mass of solution=Volume×Density=1000×2=2000 g\text{Mass of solution} = \text{Volume} \times \text{Density} = 1000 \times 2 = 2000 \, \text{g}

Since sea water is 6 M6 \, \text{M} in NaCl, 1000 mL1000 \, \text{mL} contains 66 mol of NaCl.

Mass of NaCl present is

6×58.5=351 g6 \times 58.5 = 351 \, \text{g}

Using ppm relation for dissolved O$$$_2$$,

ppm=mass of O22000×106\text{ppm} = \frac{\text{mass of O}_2}{2000} \times 10^6

So mass of O$$$_2$$ is taken as

mass of O2=1.16×10−2 g\text{mass of O}_2 = 1.16 \times 10^{-2} \, \text{g}

Now molality of O$$$_2$$ is

Molality of O2=1.16×10−2/32(2000−6×58.5)×1000\text{Molality of O}_2 = \frac{1.16 \times 10^{-2} / 32}{(2000 - 6 \times 58.5)} \times 1000

On simplifying,

=1.16×10−232×1649= \frac{1.16 \times 10^{-2}}{32 \times 1649} =0.000219=2.19×10−4= 0.000219 = 2.19 \times 10^{-4}

Thus, in the form asked, x=2x = 2.

Therefore, the required nearest integer is 22.

Detailed Working from Alternate Approach

Given: Sea water is 6 M6 \, \text{M} in NaCl, density is 2 g mL−12 \, \text{g mL}^{-1}, and dissolved oxygen concentration is 5.8 ppm5.8 \, \text{ppm}.

Find: The value of xx in the required expression.

In 1 L1 \, \text{L} of sea water, moles of NaCl are 66.

So mass of NaCl is

6×58.5=351 g6 \times 58.5 = 351 \, \text{g}

Mass of 1 L1 \, \text{L} sea water is

2×1000=2000 g2 \times 1000 = 2000 \, \text{g}

The alternate solution text uses 5.8 ppm5.8 \, \text{ppm} to write mass of dissolved O$$$_2$$ as

5.8×10−3 g5.8 \times 10^{-3} \, \text{g}

Then moles of O$$$_2$$ are

5.8×10−332=1.81×10−4\frac{5.8 \times 10^{-3}}{32} = 1.81 \times 10^{-4}

The extracted solution then reports the concentration as

2.19×10−42.19 \times 10^{-4}

and concludes that this corresponds to x=2x = 2 in the asked form.

Therefore, the correct answer is 22.

Note: The two extracted approaches contain an internal numerical inconsistency in the intermediate ppm conversion, but both source solutions conclude with x=2x = 2, and the page marks 22 as the correct answer.

Common mistakes

  • Using ppm directly as moles per million is incorrect. Here ppm is a mass-based quantity, so first convert it to mass of dissolved O$$$_2$$, then convert that mass to moles using molar mass.

  • Taking the mass of 1 L1 \, \text{L} of sea water as 1000 g1000 \, \text{g} is wrong because the density is 2 g mL−12 \, \text{g mL}^{-1}. Therefore 1000 mL1000 \, \text{mL} of solution has mass 2000 g2000 \, \text{g}.

  • Using the mass of the whole solution instead of the mass of solvent while calculating molality is incorrect. For molality, subtract the mass of dissolved NaCl from total solution mass to get the solvent mass.

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