A drug becomes ineffective after 50Take -log1054-0-097

K=2.303tlog _10 ( aa x ) k=2.30320 log_10 ( 54 ) = 0.011 month^ 1 Expiry time 't' may be calculated as : k=2.303tlog_10aa x 0.011=2.303tlog_1010050 t = 62.12 ...

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Standard VII

Chemistry

Zero Order Reaction

A drug become...

Question

A drug becomes ineffective after 50% decomposition. The original concentration of a sample was $5 m g / m L$ , which becomes $4 m g / m L$ in 20 months. Assuming the decomposition to be of first order, calculate the time in which the drug becomes inffective
Take :
${log}_{10} (\frac{5}{4}) = 0.097$

40.41 m o n t h s

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122.44 m o n t h s

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62.12 m o n t h s

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None of the above

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Solution

The correct option is C $62.12 m o n t h s$
$k = \frac{2.303}{t} l o g_{10} (\frac{a}{a - x}) k = \frac{2.303}{20} l o g_{10} (\frac{5}{4})$
$= 0.011 {month}^{- 1}$
Expiry time 't' may be calculated as :
$k = \frac{2.303}{t} l o g_{10} \frac{a}{a - x}$
$0.011 = \frac{2.303}{t} l o g_{10} \frac{100}{50}$
$t = 62.12 m o n t h s$

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A drug becomes ineffective after 50% decomposition. The original concentration of a sample was 5 mg/mL, which becomes 4 mg/mL in 20 months. Assuming the decomposition to be of first order, calculate the time in which the drug becomes inffective Take : log10(54)=0.097

The correct option is C 62.12 monthsk=2.303tlog10(aa−x)k=2.30320log10(54) =0.011 month−1 Expiry time 't' may be calculated as : k=2.303tlog10aa−x 0.011=2.303tlog1010050 t=62.12 months

A drug becomes ineffective after 50% decomposition. The original concentration of a sample was $5 m g / m L$ , which becomes $4 m g / m L$ in 20 months. Assuming the decomposition to be of first order, calculate the time in which the drug becomes inffective
Take :
${log}_{10} (\frac{5}{4}) = 0.097$