Among $$10^{-9}$$ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt

The number of atoms present in a given sample is obtained from the relation
$$\text{Number of atoms} = \frac{\text{mass of sample}}{\text{molar mass}} \times N_A$$
where $$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$$ is Avogadro’s constant.

In this question every element is taken in the same mass, $$m = 10^{-9}\, \text{g}$$. Since $$N_A$$ and $$m$$ are common to all, the factor that decides the number of atoms is only the molar mass $$M$$. Smaller $$M$$ ⇒ larger $$\dfrac{m}{M}$$ ⇒ larger number of atoms.

Approximate molar masses of the given elements are:
$$M_{\text{Pb}} \approx 207\ \text{g mol}^{-1}$$
$$M_{\text{Po}} \approx 209\ \text{g mol}^{-1}$$
$$M_{\text{Pr}} \approx 141\ \text{g mol}^{-1}$$
$$M_{\text{Pt}} \approx 195\ \text{g mol}^{-1}$$

Comparing their molar masses:
$$141\lt 195\lt 207\lt 209$$

Praseodymium (Pr) has the smallest molar mass, so for the same sample mass $$10^{-9}\, \text{g}$$ it provides the largest number of atoms.

Therefore, the element with the highest number of atoms is Praseodymium (Pr).
Answer: Option B (Pr)