An element has a bcc structure with a cell edge of `288` pm. The density of the element is `7.2 g cm^(-3)`. How many atoms are present in `208 g` of the element?

Explanation

To solve the problem step by step, we will follow the outlined approach based on the information provided in the question.

### Step 1: Convert the cell edge from picometers to centimeters
The given cell edge \( a \) is \( 288 \) pm. We need to convert this to centimeters.

\[ a = 288 \, \text{pm} = 288 \times 10^{-12} \, \text{m} = 288 \times 10^{-10} \, \text{cm} \]
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