Answer:
a. 6.15 mL b. 30.73 mL
Explanation:
a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?
Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.
So concentration of ketamine C = mass of ketamine, m/volume of water, V
m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL
So, C = m/V = 0.250 g/59.125 mL = 0.00423 g/mL = 4.23 mg/mL
Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg
Since mass, M = concentration ,C × volume, V
M = CV
V = M/C
The volume of ketamine required for the 0.400 mg/kg high is
V = 26 mg/4.23 mg/mL
V = 6.15 mL
b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?
Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg
Since mass, M' = concentration ,C × volume, V
M' = CV
V = M/C
The volume of ketamine required for the 2.00 mg/kg unconscious injection is
V = 130 mg/4.23 mg/mL
V = 30.73 mL
