g is known to be ineffective after it has decomposed 25% the original concentration of a sample was 500 unkits/ml when analyzed 20 months later the concentration was found to be 450 units /ml assuming that decomposition is of 1 st order what will be the expiry time of the drug?

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BIKRAM · Answer

ln(A 0 /A)= Kt : K=(1/20×ln(500/450)=5.268×10 -3 month -1 A= 500-500×(25/100)=375 units/ml expiry date (t)= (1/K)ln(500/375)= 54.6 month

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g is known to be ineffective after it has decomposed 25% the original concentration of a sample was 500 unkits/ml when analyzed 20 months later the concentration was found to be 450 units /ml assuming that decomposition is of 1 st order what will be the expiry time of the drug?

g is known to be ineffective after it has decomposed 25% the original concentration of a sample was 500 unkits/ml when analyzed 20 months later the concentration was found to be 450 units /ml assuming that decomposition is of 1^st order what will be the expiry time of the drug?

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g is known to be ineffective after it has decomposed 25% the original concentration of a sample was 500 unkits/ml when analyzed 20 months later the concentration was found to be 450 units /ml assuming that decomposition is of 1 st order what will be the expiry time of the drug?

g is known to be ineffective after it has decomposed 25% the original concentration of a sample was 500 unkits/ml when analyzed 20 months later the concentration was found to be 450 units /ml assuming that decomposition is of 1st order what will be the expiry time of the drug?

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g is known to be ineffective after it has decomposed 25% the original concentration of a sample was 500 unkits/ml when analyzed 20 months later the concentration was found to be 450 units /ml assuming that decomposition is of 1^st order what will be the expiry time of the drug?