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MCQEasyJEE 2025Mole Concept

JEE Chemistry 2025 Question with Solution

2.8×10−32.8 \times 10^{-3} mol of CO2CO_2 is left after removing 102110^{21} molecules from its ‘xx’ mg sample. The mass of CO2CO_2 taken initially is:

Given: NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23} \, mol^{-1}

  • A

    196.2 mg196.2 \, \text{mg}

  • B

    98.3 mg98.3 \, \text{mg}

  • C

    150.4 mg150.4 \, \text{mg}

  • D

    48.2 mg48.2 \, \text{mg}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: After removing 102110^{21} molecules, 2.8×10−32.8 \times 10^{-3} mol of CO2CO_2 remains.

Find: The initial mass of the CO2CO_2 sample.

First convert the removed molecules into moles using Avogadro's number:

10216.02×1023=1.66×10−3 mol\frac{10^{21}}{6.02 \times 10^{23}} = 1.66 \times 10^{-3} \, \text{mol}

So, the initial moles of CO2CO_2 were:

2.8×10−3+1.66×10−3=4.46×10−3 mol2.8 \times 10^{-3} + 1.66 \times 10^{-3} = 4.46 \times 10^{-3} \, \text{mol}

The molar mass of CO2CO_2 is 44 g/mol44 \, \text{g/mol}. Therefore, the initial mass is:

4.46×10−3 mol×44 g/mol=196.24 mg4.46 \times 10^{-3} \, \text{mol} \times 44 \, \text{g/mol} = 196.24 \, \text{mg}

Therefore, the initial mass of the CO2CO_2 sample is 196.2 mg196.2 \, \text{mg}. The correct option is A.

Molecules to moles to mass

Given: Remaining amount of CO2CO_2 is 2.8×10−32.8 \times 10^{-3} mol after removing 102110^{21} molecules.

Find: The initial mass of the sample.

Number of molecules present in the remaining 2.8×10−32.8 \times 10^{-3} mol is:

N=n×NA=2.8×10−3×6.02×1023N = n \times N_A = 2.8 \times 10^{-3} \times 6.02 \times 10^{23} N=1.6856×1021 moleculesN = 1.6856 \times 10^{21} \text{ molecules}

Add the removed molecules to get the original number of molecules:

Ntotal=1.6856×1021+1021=2.6856×1021 moleculesN_{total} = 1.6856 \times 10^{21} + 10^{21} = 2.6856 \times 10^{21} \text{ molecules}

Convert this total into moles:

n=NtotalNA=2.6856×10216.02×1023≈4.46×10−3 moln = \frac{N_{total}}{N_A} = \frac{2.6856 \times 10^{21}}{6.02 \times 10^{23}} \approx 4.46 \times 10^{-3} \, \text{mol}

Now calculate mass using molar mass 44 g/mol44 \, \text{g/mol}:

m=n×Molar Mass=4.46×10−3×44×103 mgm = n \times \text{Molar Mass} = 4.46 \times 10^{-3} \times 44 \times 10^3 \, \text{mg} m≈196.2 mgm \approx 196.2 \, \text{mg}

Thus, the initial mass of CO2CO_2 taken was 196.2 mg196.2 \, \text{mg}.

Common mistakes

  • Converting 102110^{21} molecules directly into mass without first converting to moles is incorrect. Molecules must be related to moles using NAN_A. First find moles, then multiply by molar mass.

  • Using only the remaining 2.8×10−32.8 \times 10^{-3} mol to calculate the initial mass is wrong because some molecules were removed. Add the removed amount in moles to the remaining moles before finding mass.

  • Forgetting to convert grams into milligrams can lead to a wrong option. After calculating mass in grams, convert properly using 1 g=103 mg1 \, \text{g} = 10^3 \, \text{mg}.

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