Among $ 10^{-10} $ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt
Among $ 10^{-10} $ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt
Show Hint
- Po
- Pr
- Pb
- Pt
The Correct Option is B
Approach Solution - 1
To determine which element has the highest number of atoms when each has a mass of \(10^{-10}\) grams, we must consider the concept of moles and Avogadro's number. The key is to calculate the number of moles of each element since the number of atoms is directly related to the number of moles.
The formula to calculate the number of moles is:
\(n = \frac{m}{M}\)
where:
- \(n\) = number of moles
- \(m\) = mass of the element in grams (here, \(10^{-10}\) g)
- \(M\) = molar mass of the element in grams per mole
Let's calculate the moles for each element:
- Lead (Pb): Molar mass = 207 g/mol
- Polonium (Po): Molar mass = 209 g/mol
- Praseodymium (Pr): Molar mass = 140 g/mol
- Platinum (Pt): Molar mass = 195 g/mol
The number of moles for each element is calculated as follows:
- \(n_{\text{Pb}} = \frac{10^{-10}}{207}\) moles
- \(n_{\text{Po}} = \frac{10^{-10}}{209}\) moles
- \(n_{\text{Pr}} = \frac{10^{-10}}{140}\) moles
- \(n_{\text{Pt}} = \frac{10^{-10}}{195}\) moles
Now, we compare these moles to determine which is the greatest:
- The smallest denominator (molar mass) will give the largest number of moles for the given mass.
- Pr (Praseodymium) has the smallest molar mass, hence has the highest number of moles and therefore the highest number of atoms.
Conclusion: The element with the highest number of atoms in \(10^{-10}\) grams is Praseodymium (Pr).
Approach Solution -2
$\text{No. of atoms} = \frac{\text{Mass in g}}{\text{Molar Mass (g/mol)}} \times N_A$
Therefore for the same Mass element having the least Molar mass will have the higher no. of atoms.
$M_{Pb} = 209$
$M_{Pr} = 141$
$M_{Po} = 207$
$M_{Pt} = 195$
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