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NEET CHEMISTRYMedium

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 L−1×0.025 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+HCl→NaCl+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 = 0.025−0.01875=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}.
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