Drop-on-demand (DoD) technology is an emerging form of drug delivery in which a

Question

Drop-on-demand (DoD) technology is an emerging form of drug delivery in which a reservoir is filled with a solution of an active pharmaceutical ingredient (API) dissolved in a volatile liquid, and a device sprays nanometer-scale drops of the solution onto an edible substrate, such as a small strip the size of a stick of chewing gum. The liquid evaporates very rapidly, causing the API to crystallize on the substrate. The exact dose required by a patient can be administered based on the known concentration of the API in the reservoir and the volume of solution deposited on the substrate, enabling greater dosage accuracy than can be provided by administering fractions of tablets.

(a) A DoD device is charged with a 1.20 molar solution of ibuprofen (the API) in n-hexane. The molecular weight of ibuprofen is 206.3 g/mol. If a prescribed dosage is 5.0 mg ibuprofen/kg patient weight, how many milliliters of solution should be sprayed for a 245-pound man and a 65-pound child? How many drops are in each dose, assuming that each drop is a sphere with a radius of 1 nm?

(b) The DoD device is to be automated, so that the operator enters a patient’s body weight into a computer that determines the required solution volume and causes that volume to be sprayed on the substrate. Derive a formula for the volume, Vdose (mL), in terms of the following variables: Ms (mol API/L) = molarity of reservoir solution SGs = specific gravity of reservoir solution MWAP I (g/mol) = molecular weight of API D (mg AOI/kg body weight) = prescribed dosage WP (lbf ) = patient’s weight Check your formula by verifying your solution to Part (a).

(c) Calculate the surface-to-volume ratio of a sphere of radius r. Then calculate the total drop surface area of 1 mL (= 1 cm3 ) of the solution if it were sprayed as drops of (i) radius 1 nm and (ii) 1 mm. Speculate on the likely reason for spraying nanoscale drops instead of much larger drops.

Explanation / Answer

245lB=111.3Kg , 65 pound =29.48Kg

Dose required: for 245 pound man = 5mg* 111.3 = 556.6 mg

for child = 5mg* 29.48 = 147.4 mg

No of moles required: for adult .5566/206.3 = .00269moles

No of moles required : for child .1474/206.3 = .00071moles

In one litre, 1.2 moles is present.

.00269 moles is present in: .00269/1.2 = 2.24mL

.00071 moles is present in: .00071/1.2= .59mL

b) The formula is: Vdose(mL):(D*WP*.453)/(MWAPI*d*Ms*1000)

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