. Shake a quantity of cream (0.1564 g) containing about 7.5 mg of acyclovir with

Question

. Shake a quantity of cream (0.1564 g) containing about 7.5 mg of acyclovir with 50 mL of 0.5 M sulfuric acid. Shake well with 50 mL of ethyl acetate, allow to separate and collect the lower aqueous phase. Wash the organic layer with 20 mL of 0.5 M sulfuric acid and dilute the combined washings and the aqueous layer to 100 mL with 0.5 M sulfuric acid. Mix well and filter. Discard the first few mLs of filtrate and to 10.00 mL of the filtrate add water to produce 50.00 mL. Measure the absorbance of the resulting solution at the maximum at 255 nm. Using the data below calculate the % w/w of cyclovir in the cream and answer this question: Why was the cream shaken with ethyl acetate if the aqueous layer was measured?

Absorbance reading of diluted sampled = 0.815

A 1%w/v standard solution using 1cm cell at 255 nm gave the constant “a” a value of 562 in the equation A = abc.

Explanation / Answer

Absrobance = a X concentration x length

0.815 = 562 X concentration X 1cm

Concentration = 689.57

The cream was shaken with ethylacetate to remove the organic components present in it. The ethyl acetate is used to separate the unreacted acyclovir from the cream.

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