1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution. The mass of sodium hydroxide left unreacted is equal to: - Chemistry MCQ NEET (UG) 2024 2026 | ExamTest.live

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1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution. The mass of sodium hydroxide left unreacted is equal to:

Solution & Step-by-step Explanation

Reaction: NaOH + HCl \to NaCl + H_{2} O Moles of NaOH initially = \frac{Mass}{Molar Mass} = \frac{1}{40} = 0.025 mol .Moles of HCl used = Molarity \times Volume (L) = 0.75 \times 0.025 = 0.01875 mol .Since the stoichiometry is 1 : 1, 0.01875 mol of HCl will react with 0.01875 mol of NaOH .Moles of NaOH unreacted = 0.025 - 0.01875 = 0.00625 mol .Mass of NaOH unreacted = 0.00625 \times 40 = 0.25 g = 250 mg .

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1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution. The mass of sodium hydroxide left unreacted is equal to:

250 mg

Zero mg

200 mg

750 mg

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