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A preparation of drug granules weighing 2 g with a total surface area of (0.35 × 104 cm2) is allowed to dissolve in 250 mL water at 25 °C. After the first minute, 0.25 g has passed into solution. If the solubility (Cs) of the drug is 20 mg/mL at 25 °C, calculate the dissolution rate

a.

4.2 mg/min

b.

4.2 mg/sec

c.

4.2 mg/mL

d.

250 mg/sec

A Preparation Of Drug Granules Weighing 2 G With A Total Surface Area Of 035 104 Cm2 Is Allowed To Dissolve In 250 ML Water At 25 C After The First Minute 025 G class=

Sagot :

Explanation:

The Noyes-Whitney equation is often used to predict the rate at which solid substances, such as drug tablets and granules, in a solvent. Hence, according to the Noyes-Whitney equation, the dissolution rate of a drug tablet or granule can be evaluated in accordance with the following expression

                                              [tex]\displaystyle\frac{dm}{dt} \ = \ A \ \displaystyle\frac{D}{h}(C_{s} - C_{b})[/tex],

where [tex]\displaystyle\frac{dm}{dt}[/tex] is the rate of dissolution, [tex]D[/tex] is the dissociation coefficient, [tex]h[/tex] is the thickness of the diffusion layer, [tex]A[/tex] is the surface area of the solute particle, [tex]C_{s}[/tex] is the particle surface (saturation) concentration and [tex]C_{b}[/tex] is the concentration in the bulk solvent.

In the case of the question, we need to find the dissolution rate of the drug granules which corresponds to the first-order differential term, [tex]\displaystyle\frac{dm}{dt}[/tex]. Looking closely at this term, it implies that the dissolution rate is defined by the change of the solute mass, [tex]\Delta m[/tex], with respect to the change in time, [tex]\Delta t[/tex].

In accordance with the given description from the question, we know that 0.25g of the 2g drug granules had dissolved into the solution over a period of 1 minute. Using dimensional analysis, we can simply calculate the dissolution rate as

                                        [tex]\displaystyle\frac{dm}{dt} \ = \ \displaystyle\frac{\Delta m}{\Delta t} \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{(0.25 \times 10^3) \ \text{mg}}{60 \ \text{sec}} \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{250 \ \text{mg}}{60 \ \text{sec}} \\ \\ \-\hspace{0.7cm} = \ 4.2 \ \text{mg sec}^{-1} \ \ \ \ \ (1 \ \text{d.p.})[/tex]