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
- 750 mg
- 250 mg
- Zero mg
- 200 mg
Answer
(2) 250 mg
Solution
Moles of NaOH = 140 = 0.025 mol = 25 mmol
Moles of HCl = 25 0.75 = 18.75 mmol
NaOH + HCl NaCl + H2O (1 : 1)
NaOH left = 25 - 18.75 = 6.25 mmol
Mass left = 6.25 40 = 250 mg
Answer: (2)