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NEET CHEMISTRYSome Basic Concepts of ChemistryMedium

Question

If 1 gram1 \text{ gram} of sodium hydroxide was treated with 25 mL25 \text{ mL} of 0.75 M0.75 \text{ M} HCl solution, the mass of sodium hydroxide left unreacted is equal to:

A

250 mg

B

Zero mg

C

200 mg

D

750 mg

Step-by-Step Solution

  1. Calculate Moles of Reactants:
  • Moles of NaOH (nNaOHn_{\text{NaOH}}) = MassMolar Mass=1 g40 g/mol=0.025 mol\frac{\text{Mass}}{\text{Molar Mass}} = \frac{1 \text{ g}}{40 \text{ g/mol}} = 0.025 \text{ mol}.
  • Moles of HCl (nHCln_{\text{HCl}}) = Molarity×Volume (L)\text{Molarity} \times \text{Volume (L)} [Class 12 Chemistry, Eq. 1.8]. nHCl=0.75 mol L1×0.025 L=0.01875 moln_{\text{HCl}} = 0.75 \text{ mol L}^{-1} \times 0.025 \text{ L} = 0.01875 \text{ mol}
  1. Determine Limiting Reagent:
  • The balanced reaction is: NaOH+HClNaCl+H2O\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}.
  • Stoichiometry is 1:1. Since 0.01875 mol<0.025 mol0.01875 \text{ mol} < 0.025 \text{ mol}, HCl is the limiting reagent.
  1. Calculate Unreacted NaOH:
  • Moles consumed = Moles of HCl = 0.01875 mol0.01875 \text{ mol}.
  • Moles remaining = 0.0250.01875=0.00625 mol0.025 - 0.01875 = 0.00625 \text{ mol}.
  1. Convert to Mass:
  • Mass = Moles×Molar Mass=0.00625 mol×40 g/mol=0.25 g\text{Moles} \times \text{Molar Mass} = 0.00625 \text{ mol} \times 40 \text{ g/mol} = 0.25 \text{ g}.
  • Convert to mg: 0.25 g×1000 mg/g=250 mg0.25 \text{ g} \times 1000 \text{ mg/g} = 250 \text{ mg}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Some Basic Concepts of Chemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSome Basic Concepts of Chemistrysodiumhydroxidetreatedsolutionsodium

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