KVPY Sample Paper KVPY Stream-SX Model Paper-24

An organic compound undergoes first order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are \[{{t}_{1/8}}\] and \[{{t}_{1/10}}\] respectively. What is the value of \[\frac{\left[ {{t}_{1/8}} \right]}{\left[ {{t}_{1/10}} \right]}\times 10?\]\[({{\log }_{10}}2=0.3)\]

A) 4

B) 8

C) 9

D) 11

Correct Answer: C

Solution :

For a first order process,\[kt=ln\frac{{{[A]}_{0}}}{\left[ A \right]}\] where, \[{{[A]}_{0}}\] = initial concentration.
[A] = concentration of reactant remaining at time
\[k{{t}_{1/8}}=ln\frac{{{\left[ A \right]}_{0}}}{{{\left[ A \right]}_{0}}/8}=ln\,8\]	?.. (i)
\[k{{t}_{1/10}}=ln\frac{{{\left[ A \right]}_{0}}}{{{\left[ A \right]}_{0}}/10}=ln\,10\]	??(ii)
Therefore,
\[\frac{{{t}_{1/8}}}{{{t}_{1/10}}}=\frac{ln\,\,8}{ln\,\,10}=\log 8=\log {{2}^{3}}3\log 2\]
\[\frac{{{t}_{1/8}}}{{{t}_{1/10}}}=3\times 0.3=0.9\]
\[\frac{{{t}_{1/8}}}{{{t}_{1/10}}}=10=0.9\times 10=0.9\]