Question 8
Volumetric Analysis3.0 g of a mixture of potassium carbonate and potassium chloride were dissolved in a 250 cm3 standard flask. 25cm3 of this solution required 40.00 cm3 of 0.1 M HCl for neutralization. What is the percentage by weight of K2CO3 in the mixture (K = 39, O = 16, C = 12).
- A.60
- B.72
- C.82
- D.89
- E.92
Show the answer and explanation
Answer: E (92)
Explanation
HCl used = 0.040 × 0.1 = 0.004 mol, which reacts with 0.002 mol of K2CO3 (K2CO3 + 2HCl). The whole 250 cm3 holds 0.02 mol × 138 g = 2.76 g, and 2.76/3.0 × 100 = 92%.