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Redox Reactions question

2019 · 12 Jan · Shift 1 · Q14
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Redox Reactions question

2019 · 12 Jan · Shift 1 · Q14

JEE MainChemistryRedox ReactionsMCQ+4 / −1
50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is -
  1. A
    20 g
  2. B
    4 g
  3. C
    80 g
  4. D
    10 g
View written solutionFree

Correct answer: B

  1. Write the balanced neutralization reaction

Oxalic acid is a dibasic acid: H2C2O4+2NaOH→Na2C2O4+2H2O\mathrm{H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O}H2​C2​O4​+2NaOH→Na2​C2​O4​+2H2​O

So,

  • 111 mole oxalic acid reacts with 222 moles NaOH.
  1. Calculate moles of oxalic acid used

Given:

  • Volume of oxalic acid =50 mL=0.050 L= 50\,\text{mL} = 0.050\,\text{L}=50mL=0.050L
  • Molarity of oxalic acid =0.5 M= 0.5\,\text{M}=0.5M

Moles of oxalic acid: n=M×V=0.5×0.050=0.025 moln = M \times V = 0.5 \times 0.050 = 0.025\,\text{mol}n=M×V=0.5×0.050=0.025mol

  1. Find moles of NaOH neutralized

From stoichiometry: 1 mol oxalic acid:2 mol NaOH1\,\text{mol oxalic acid} : 2\,\text{mol NaOH}1mol oxalic acid:2mol NaOH

Therefore, moles of NaOH=2×0.025=0.050 mol\text{moles of NaOH} = 2 \times 0.025 = 0.050\,\text{mol}moles of NaOH=2×0.025=0.050mol

This amount of NaOH is present in 25 mL25\,\text{mL}25mL of the NaOH solution.

  1. Find moles of NaOH in 50 mL of the same solution

Since 50 mL50\,\text{mL}50mL is double of 25 mL25\,\text{mL}25mL, moles in 50 mL=2×0.050=0.100 mol\text{moles in 50 mL} = 2 \times 0.050 = 0.100\,\text{mol}moles in 50 mL=2×0.050=0.100mol

  1. Convert moles to mass

Molar mass of NaOH: 40 g mol−140\,\text{g mol}^{-1}40g mol−1

Mass of 0.1000.1000.100 mol NaOH: m=n×M=0.100×40=4 gm = n \times M = 0.100 \times 40 = 4\,\text{g}m=n×M=0.100×40=4g

  1. Match with options

4 g\boxed{4\,\text{g}}4g​

So the correct option is B.

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