A) 32.4 mm
B) 51.2 mm
C) 46.8 mm
D) 24.1 min
Correct Answer: D
Solution :
[d] \[k=\frac{2.303}{t}\log \left( \frac{a}{a-x} \right)\] \[\frac{kt}{2.303}=\log a-\log (a-x)\] \[\frac{k{{t}_{1}}}{2.303}=\log a-\log (a-{{x}_{1}})at\,time\,{{t}_{1}}\] ...(i) \[\frac{k{{t}_{2}}}{2.303}=\log a-\log (a-{{x}_{2}})at\,time\,{{t}_{2}}\] ...(ii) Subtracting (i) and (ii) \[\therefore \frac{k}{2.303}({{t}_{2}}-{{t}_{1}})=log\left( \frac{a-{{x}_{1}}}{a-{{x}_{2}}} \right)\] Rate, \[{{r}_{1}}=k(a-{{x}_{1}});{{r}_{2}}=k(a-{{x}_{2}})\] \[\therefore \frac{(a-{{x}_{1}})}{(a-{{x}_{2}})}=\frac{{{r}_{1}}}{{{r}_{2}}}\Rightarrow \frac{0.04}{0.03}=\frac{4}{3}\] \[\frac{k(40-30)}{2.303}=\log \frac{4}{3}\] \[\therefore k=\frac{2.303}{10}\log \frac{4}{3}=\frac{2.303}{10}\log 1.33\] \[=\frac{2.303\times 0.125}{10}=0.02878{{\min }^{-1}}\] \[{{T}_{50}}=\frac{0.693}{0.02878}=24.1\min \]
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