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**Hint:**We are given the rate constant of a second order reaction to be ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$. We will try to find the relation and convert moles into a number of molecules, litres into cubic centimetre and second into minutes. And finally we will put it in ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$ to find the required answer.

**Complete step by step answer:**

As the rate constant of a second order reaction is given ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$and we have to find the rate constant of the second order reaction in the unit $cc.{molecule^{ - 1}}.{\min ^{ - 1}}$ . We know that

$

1L = {10^3}c{m^3} \\

1mol = 6.022 \times {10^{23}} \\

1\sec = \dfrac{1}{{60}}\min \\

$

We will have to put the required units of volume , mole and time in ${10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}}$ to find the answer in $cc.{molecule^{ - 1}}.{\min ^{ - 1}}$, We will have to do it carefully so that we can avoid calculation error. So:

$

{10^{ - 2}}lit.mol{e^{ - 1}}.{\sec ^{ - 1}} \\

= \dfrac{{{{10}^{ - 2}}{{.10}^3}cc}}{{6.022 \times {{10}^{23}}molecule.\dfrac{1}{{60}}\min }} = 9.96 \times {10^{ - 22}}cc.{molecule^{ - 1}}.{\min ^{ - 1}} \\

$

From the above explanation and calculation it is clear to us that

**The correct answer of the given question is option: A. $9.96 \times {10^{ - 22}}$**

**Additional information:**

The Rate constant is a proportional constant of the equation which expresses the Relation between the rate of the chemical equation and the concentrations of the reactants of the chemical equation. One mole of a substance contains $6.022 \times {10^{23}}$particles. This is a special number and it is also known as the Avogadro’s number or Avogadro constant. It is named after the great scientist Amedeo Avogadro. Amedeo Avogadro was from Italy.

**Note:**

Always remember that one mole has $6.022 \times {10^{23}}$particles , one litre is equal to ${10^3}c{m^3}$. One minute has sixty seconds. Always try to avoid silly mistakes and calculation errors while solving the numerical. To convert a given value in it’s different unit, always use the concept of units and dimensions as used in the above question.

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