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Practical Organic Chemistry question

2025 · 29 Jan · Shift 2 · Q23
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Practical Organic Chemistry question

2025 · 29 Jan · Shift 2 · Q23

JEE MainChemistryPractical Organic ChemistryNumerical+4 / −1
In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ‾×10−1%\underline{\hspace{2cm}}\times 10^{-1} \%​×10−1%. (Molar mass : O=16, S=32,Ba=137\mathrm{O}=16, \mathrm{~S}=32, \mathrm{Ba}=137O=16, S=32,Ba=137 in g mol−1\mathrm{g} ~\mathrm{mol}^{-1}g mol−1 )
Numerical answer
View written solutionFree

Correct answer: 275

  1. Find molar mass of BaSO4\mathrm{BaSO_4}BaSO4​

M(BaSO4)=137+32+4(16)=137+32+64=233 g mol−1M(\mathrm{BaSO_4}) = 137 + 32 + 4(16) = 137 + 32 + 64 = 233\ \text{g mol}^{-1}M(BaSO4​)=137+32+4(16)=137+32+64=233 g mol−1

  1. Mass of sulphur present in 0.40 g0.40\ \text{g}0.40 g of BaSO4\mathrm{BaSO_4}BaSO4​

In 233 g233\ \text{g}233 g of BaSO4\mathrm{BaSO_4}BaSO4​, mass of sulphur =32 g=32\ \text{g}=32 g.

So in 0.40 g0.40\ \text{g}0.40 g of BaSO4\mathrm{BaSO_4}BaSO4​,

Mass of S=32233×0.40=12.8233≈0.05494 g\text{Mass of S} = \frac{32}{233}\times 0.40 = \frac{12.8}{233} \approx 0.05494\ \text{g}Mass of S=23332​×0.40=23312.8​≈0.05494 g

  1. Calculate percentage of sulphur in the organic compound

Mass of organic compound =0.20 g=0.20\ \text{g}=0.20 g

%S=0.054940.20×100\%\text{S} = \frac{0.05494}{0.20}\times 100%S=0.200.05494​×100

%S≈27.47%\%\text{S} \approx 27.47\%%S≈27.47%

  1. The question asks in the form:

‾×10−1%\underline{\hspace{2cm}}\times 10^{-1}\%​×10−1%

Let the blank be xxx. Then

x×10−1%=27.47%x\times 10^{-1}\% = 27.47\%x×10−1%=27.47%

So,

x=27.47×10=274.7≈275x = 27.47\times 10 = 274.7 \approx 275x=27.47×10=274.7≈275

  1. Final integer answer

275\boxed{275}275​

  1. Comparison with stored correct answer

Stored correct answer = 275275275

This matches the derived answer.

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