Question

When selling on the street, dealers may not know the purity of the ketamine they have, and thus users do not know exactly how much ketamine they are receiving. It is unlikely that the ketamine is pure, or even that different batches of ketamine have the same purity. Assume the drug the user typically buys is only 25% ketamine, and therefore, the user actually dissolved 0.250 g ketamine in 1/4 cup of water to make the solution instead of 1 g in the previous question. 1 cup = 236.5 mL What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg? volume: mL What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg? of use contact us help What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?

Answer 1

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