0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm³. If 15.0 cm³ of the solution requires 12.50 cm³ of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution

0.30 mol dm³

0.40 mol dm³

0.50 mol dm³

0.60 mol dm³

Correct answer is A

M₁ V₁ = M₂ V₂
∴ M_n V_n represent volume and molar concentration of HCL also MyVy represent volume and molar concentration of Na₂CO₃
M_nV_n = MyVy
0.25 * 15 = 12.50 * My
My = (0.25 * 15)/12.50
My = 0.3m/dm³

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0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm³. If 15.0 cm³ of the solution requires 12.50 cm³ of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution

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0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm3. If 15.0 cm3 of the solution requires 12.50 cm3 of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution

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0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm³. If 15.0 cm³ of the solution requires 12.50 cm³ of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution