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Some Basic Concepts of Chemistry question

2022 · 28 Jul · Shift 2 · Q14
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Some Basic Concepts of Chemistry question

2022 · 28 Jul · Shift 2 · Q14

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
2L of 0.2M H2SO4H_2SO_4H2​SO4​ is reacted with 2L of 0.1M NaOHNaOHNaOH solution, the molarity of the resulting product Na2SO4Na_2SO_4Na2​SO4​ in the solution is ‾\underline{\hspace{2cm}}​ millimolar. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Write the balanced reaction

H2SO4+2NaOH→Na2SO4+2H2OH_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2OH2​SO4​+2NaOH→Na2​SO4​+2H2​O

  1. Calculate moles of reactants

For sulfuric acid: n(H2SO4)=M×V=0.2×2=0.4 moln(H_2SO_4)=M\times V=0.2\times 2=0.4\text{ mol}n(H2​SO4​)=M×V=0.2×2=0.4 mol

For sodium hydroxide: n(NaOH)=M×V=0.1×2=0.2 moln(NaOH)=M\times V=0.1\times 2=0.2\text{ mol}n(NaOH)=M×V=0.1×2=0.2 mol

  1. Find the limiting reagent

From the balanced equation: 1 mol H2SO4 requires 2 mol NaOH1\text{ mol }H_2SO_4 \text{ requires } 2\text{ mol }NaOH1 mol H2​SO4​ requires 2 mol NaOH

To completely react 0.40.40.4 mol H2SO4H_2SO_4H2​SO4​, required NaOH would be: 0.4×2=0.8 mol0.4\times 2=0.8\text{ mol}0.4×2=0.8 mol

But available NaOH is only 0.20.20.2 mol, so NaOH is the limiting reagent.

  1. Calculate moles of }Na_2SO_4\text{ formed}

According to stoichiometry: 2 mol NaOH→1 mol Na2SO42\text{ mol }NaOH \rightarrow 1\text{ mol }Na_2SO_42 mol NaOH→1 mol Na2​SO4​

Hence, n(Na2SO4)=0.22=0.1 moln(Na_2SO_4)=\frac{0.2}{2}=0.1\text{ mol}n(Na2​SO4​)=20.2​=0.1 mol

  1. Calculate total volume of solution

Assuming volumes are additive: Vtotal=2+2=4 LV_{\text{total}}=2+2=4\text{ L}Vtotal​=2+2=4 L

  1. **Calculate molarity of }Na_2SO_4$$

M(Na2SO4)=0.14=0.025 MM(Na_2SO_4)=\frac{0.1}{4}=0.025\text{ M}M(Na2​SO4​)=40.1​=0.025 M

Convert to millimolar: 0.025 M=25 mM0.025\text{ M}=25\text{ mM}0.025 M=25 mM

  1. Final answer

25\boxed{25}25​

This matches the stored correct answer.

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