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Some Basic Concepts of Chemistry question

2021 · 16 Mar · Shift 2 · Q21
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Some Basic Concepts of Chemistry question

2021 · 16 Mar · Shift 2 · Q21

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, ‾×\underline{\hspace{2cm}}\times​× 10 −-− 5 moles of lead sulphate precipitate out. (Round off to the Nearest Integer).
Numerical answer
View written solutionFree

Correct answer: 525

  1. Write the reaction involved

Lead nitrate reacts with chromic sulphate by double displacement:

3Pb(NO3)2+Cr2(SO4)3→3PbSO4↓+2Cr(NO3)33\text{Pb}(\text{NO}_3)_2 + \text{Cr}_2(\text{SO}_4)_3 \rightarrow 3\text{PbSO}_4 \downarrow + 2\text{Cr}(\text{NO}_3)_33Pb(NO3​)2​+Cr2​(SO4​)3​→3PbSO4​↓+2Cr(NO3​)3​

Thus,

  • 333 mol Pb(NO3)2\text{Pb}(\text{NO}_3)_2Pb(NO3​)2​ give 333 mol PbSO4\text{PbSO}_4PbSO4​
  • so effectively, 111 mol Pb2+\text{Pb}^{2+}Pb2+ gives 111 mol PbSO4\text{PbSO}_4PbSO4​
  • 111 mol Cr2(SO4)3\text{Cr}_2(\text{SO}_4)_3Cr2​(SO4​)3​ provides 333 mol SO42−\text{SO}_4^{2-}SO42−​
  1. Calculate moles of lead nitrate

Volume =35 mL=0.035 L= 35\,\text{mL} = 0.035\,\text{L}=35mL=0.035L

Molarity =0.15 M= 0.15\,\text{M}=0.15M

n(Pb(NO3)2)=M×V=0.15×0.035=0.00525 moln\big(\text{Pb}(\text{NO}_3)_2\big)= M \times V = 0.15 \times 0.035 = 0.00525\,\text{mol}n(Pb(NO3​)2​)=M×V=0.15×0.035=0.00525mol

So moles of Pb2+\text{Pb}^{2+}Pb2+ available =0.00525= 0.00525=0.00525 mol.

  1. Calculate moles of chromic sulphate

Volume =20 mL=0.020 L= 20\,\text{mL} = 0.020\,\text{L}=20mL=0.020L

Molarity =0.12 M= 0.12\,\text{M}=0.12M

n(Cr2(SO4)3)=0.12×0.020=0.0024 moln\big(\text{Cr}_2(\text{SO}_4)_3\big)=0.12 \times 0.020 = 0.0024\,\text{mol}n(Cr2​(SO4​)3​)=0.12×0.020=0.0024mol

Each mole of Cr2(SO4)3\text{Cr}_2(\text{SO}_4)_3Cr2​(SO4​)3​ gives 333 mol SO42−\text{SO}_4^{2-}SO42−​, so

n(SO42−)=3×0.0024=0.0072 moln(\text{SO}_4^{2-}) = 3 \times 0.0024 = 0.0072\,\text{mol}n(SO42−​)=3×0.0024=0.0072mol

  1. Find the limiting reagent

Reaction for precipitate formation:

Pb2++SO42−→PbSO4↓\text{Pb}^{2+} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 \downarrowPb2++SO42−​→PbSO4​↓

Available:

  • Pb2+=0.00525\text{Pb}^{2+} = 0.00525Pb2+=0.00525 mol
  • SO42−=0.0072\text{SO}_4^{2-} = 0.0072SO42−​=0.0072 mol

Since 0.00525<0.00720.00525 < 0.00720.00525<0.0072, Pb2+\text{Pb}^{2+}Pb2+ is the limiting reagent.

Hence moles of PbSO4\text{PbSO}_4PbSO4​ precipitated:

n(PbSO4)=0.00525 moln(\text{PbSO}_4)=0.00525\,\text{mol}n(PbSO4​)=0.00525mol

  1. Express in the required form

We need:

‾×10−5 mol\underline{\hspace{2cm}} \times 10^{-5}\text{ mol}​×10−5 mol

Now,

0.00525=5.25×10−3=525×10−50.00525 = 5.25 \times 10^{-3} = 525 \times 10^{-5}0.00525=5.25×10−3=525×10−5

So the required integer is

525\boxed{525}525​

  1. Comparison with stored answer

Stored correct answer = 525525525

Our derived answer = 525525525

So the answer matches.

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