Question:

1 gram of sodium hydroxide was treated with 25 ml. of 0.75 M HCI solution, the mass of sodium hydroxide left unreacted is equal to

Updated On: Jul 2, 2025
  • 750mg
  • 250 mg
  • Zero mg
  • 200 mg
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The Correct Option is B

Solution and Explanation

Step 1: Write the Reaction 

The neutralization reaction between NaOH and HCl is:

$$ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + H_2O $$

Step 2: Calculate the Moles of NaOH

Given:

  • Mass of NaOH = 1 g
  • Molar mass of NaOH = 40 g/mol

Moles of NaOH:

$$ \text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1}{40} = 0.025 \text{ mol} $$

Step 3: Calculate the Moles of HCl

Given:

  • Volume of HCl = 25 mL = 0.025 L
  • Molarity of HCl = 0.75 M

Moles of HCl:

$$ \text{Moles of HCl} = M \times V = 0.75 \times 0.025 = 0.01875 \text{ mol} $$

Step 4: Determine the Limiting Reagent

The reaction follows a 1:1 molar ratio between NaOH and HCl.

  • NaOH available: 0.025 mol
  • HCl available: 0.01875 mol

Since HCl has fewer moles, it is the limiting reagent.

Step 5: Calculate Unreacted NaOH

Unreacted moles of NaOH:

$$ \text{Unreacted NaOH} = 0.025 - 0.01875 = 0.00625 \text{ mol} $$

Mass of unreacted NaOH:

$$ \text{Mass} = 0.00625 \times 40 = 0.25 \text{ g} = 250 \text{ mg} $$

Step 6: Conclusion

The mass of unreacted NaOH is 250 mg.

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