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If 24.83 cm3 of 0.15 M NaOH is titrated to its end point with 39.45 cm3 of HCl, what is ...

If 24.83 cm3 of 0.15 M NaOH is titrated to its end point with 39.45 cm3
of HCl, what is the molarity of the HCl?
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
  • A 0.094 M
  • B 0.150 M
  • C 0.940 M
  • D 1.500 M
Correct Answer: Option A
Explanation:
NaOH + HCl → NaCl + H2O moles ratio = 1/1
↔ (MaVa) / Mb x 39.45 = 0.094

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