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0.25 mole of hydrogen chloride was dissolved in distilled water and the volume m...


Question

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

Options

A) 0.30 mol dm3

B) 0.40 mol dm3

C) 0.50 mol dm3

D) 0.60 mol dm3


The correct answer is A.

Explanation:

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

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Dicussion (1)

  • M1 V1 = M2 V2
    ∴ Mn Vn represent volume and molar concentration of HCL also MyVy represent volume and molar concentration of Na2CO3
    MnVn = MyVy
    0.25 * 15 = 12.50 * My
    My = (0.25 * 15)/12.50
    My = 0.3m/dm3

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