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Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 4 April, Evening Shift — Question 13

Sea water, which can be considered as a 6 molar ( 6 M ) solution of NaCl , has a density of 2 g mL−12 \mathrm{~g} \mathrm{~mL}{ }^{-1}. The concentration of dissolved oxygen (O2)\left(\mathrm{O}_{2}\right) in sea water is 5.8 ppm . Then the concentration of dissolved oxygen (O2)\left(\mathrm{O}_{2}\right) in sea water, is x×10−4 mx \times 10^{-4} \mathrm{~m}. x=x= _____\_\_\_\_\_ (Nearest integer) Given: Molar mass of NaCl is 58.5 g mol−158.5 \mathrm{~g} \mathrm{~mol}^{-1} Molar mass of O2\mathrm{O}_{2} is 32 g mol−132 \mathrm{~g} \mathrm{~mol}^{-1}

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Given 5.8 ppm of O2\mathrm{O}_{2}, means 5.8mgO25.8 \mathrm{mg} \mathrm{O}_{2} in 1 L of sea water

or 5.8×10−3 gO25.8 \times 10^{-3} \mathrm{~g} \mathrm{O}_{2} in 1 L sea water

number of moles of O2=5.8×10−332\mathrm{O}_{2}=\frac{5.8 \times 10^{-3}}{32} in 1 L

Molarity of O2=5.8×10−332M=1.8125×10−4M\mathrm{O}_{2}=\frac{5.8 \times 10^{-3}}{32} \mathrm{M}=1.8125 \times 10^{-4} \mathrm{M}

Since mass of solute is very less than solvent so molality = molarity

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic
Concentration Terms and Their Interconversion

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