Chemistry
JAMB 1999
0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm
3. If 15.0 cm
3 of the solution requires 12.50 cm
3 of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution
-
A.
0.30 mol dm3
-
B.
0.40 mol dm3
-
C.
0.50 mol dm3
-
D.
0.60 mol dm3
Correct Answer: Option 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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