Use the following information to answer question 12-15
Solution A is 0.1mol/dm² HCI and solution B contains 5g of Impure NaOH in one liter. The average volume of acid
used to neutralize 25cm^{3} the base was determined to be 20.50~cm^{3}
12. The concentration of B in moldm³...... (A) 0.082 mol/dm³ (B) 0.12 mol/dm³ (C) 0.10~mol/dm^{3}\{0\} 0.21 mol/dm³
13. The concentration of B in g/dm^{3}(A) 3.51g (B) 3.28g (C) 4.8 (D) 3.48
14. The percentage purity of the NaOH is (A) 70.2% (0) 65.6% (D) 69.6% (0) 96.0%
15. The mole ratio for the reaction between A and B is (A) 1:1 (6) 1:2 (C) 2:1 (D) none of these
12. We know that the volume of acid used to neutralize 25cm³ of the base (solution B) is 20.50cm³. Since the concentration of solution A is given as 0.1mol/dm², we can use the volume and concentration to calculate the moles of acid used in the reaction:
moles of acid = concentration × volume (in dm³)
moles of acid = 0.1 mol/dm² × 20.50 cm³ × (1/1000) dm³/cm³
moles of acid = 0.00205 mol
Since the reaction between acid and base is 1:1, we can assume that the moles of base used in the reaction is also 0.00205 mol.
Now, we can calculate the concentration of solution B:
concentration of B = moles of base / volume (in dm³)
concentration of B = 0.00205 mol / 0.025 dm³
concentration of B = 0.082 mol/dm³
Therefore, the concentration of B is 0.082 mol/dm³.
Answer: (A) 0.082 mol/dm³
13. The molar mass of NaOH is 23 + 16 + 1 = 40 g/mol. Since the concentration of B is 0.082 mol/dm³ and the volume is 1 dm³, we can calculate the mass of NaOH in the solution:
mass of NaOH = concentration × molar mass
mass of NaOH = 0.082 mol/dm³ × 40 g/mol
mass of NaOH = 3.28 g
Therefore, the concentration of B in g/dm³ is 3.28 g.
Answer: (B) 3.28g
14. To calculate the percentage purity of NaOH, we need to know how much of the 5g of NaOH in solution B is actually pure NaOH. Since the reaction with the acid is 1:1, the moles of acid used (0.00205 mol) will equal the moles of pure NaOH in solution B.
mass of pure NaOH = moles of pure NaOH × molar mass
mass of pure NaOH = 0.00205 mol × 40 g/mol
mass of pure NaOH = 0.082 g
The percentage purity of NaOH can then be calculated:
percentage purity = (mass of pure NaOH / mass of impure NaOH) × 100%
percentage purity = (0.082 g / 5 g) × 100%
percentage purity = 1.64%
Therefore, the percentage purity of NaOH is 1.64%.
Answer: (D) 69.6%
15. The mole ratio for the reaction between A and B is 1:1. This means that for every mole of acid (solution A) used, one mole of base (solution B) reacts. The stoichiometry of the reaction determines the mole ratio, and since the reaction is 1:1, we can conclude that the mole ratio is 1:1.
Answer: (A) 1:1