Question

which of the following has the largest number of atoms a.1gAu(s) b.1gNa(s) c.1gLi(s) d.1gcl2(g)​

Answer 1

Answer:

the largest no. of 1 gram of Li

Answer 2

Among the given elements 1g Li has largest number of atoms.

Explanation:

• To find the elements with largest number of atoms, number of atoms present in each of the given element must be calculated.
• Accordingly, the following formula can be used to find the number of atoms:

Number of atoms = $\frac{Given Mass}{Molar mass}\times 6.02\times 10^{23}$

(a) 1g of Au

Number of atoms = $\frac{Given Mass}{Molar mass}\times 6.02\times 10^{23}$

Number of atoms = $\frac{1}{197}\times 6.02\times 10^{23}$

Number of atoms = $\frac{6.02\times 10^{23}}{197}$

Number of atoms = $3.05\times10^{21}$

(b) 1g of Na

Number of atoms = $\frac{Given Mass}{Molar mass}\times 6.02\times 10^{23}$

Number of atoms = $\frac{1}{23}\times 6.02\times 10^{23}$

Number of atoms = $\frac{6.02\times 10^{23}}{23}$

Number of atoms = $26.17\times10^{21}$

(c) 1g of Li

Number of atoms = $\frac{Given Mass}{Molar mass}\times 6.02\times 10^{23}$

Number of atoms = $\frac{1}{7}\times 6.02\times 10^{23}$

Number of atoms = $\frac{6.02\times 10^{23}}{7}$

Number of atoms = $86\times10^{21}$

(d) 1g of $Cl_2$

Number of atoms = $\frac{Given Mass}{Molar mass}\times 6.02\times 10^{23}$

Number of atoms = $\frac{1}{71}\times 6.02\times 10^{23}$

Number of atoms = $\frac{6.02\times 10^{23}}{71}$

Number of atoms = $8.47\times10^{21}$

From the above calculation the following data is obtained:

• 1g of Au contains $3.05\times10^{21}$ atoms.

• 1g of Na contains $26.17\times10^{21}$ atoms.

• 1g of Li contains $86\times10^{21}$ atoms.

• 1g of $Cl_2$ contains $8.47\times10^{21}$ atoms.

From the above values it can be inferred that 1g of Li has largest number of atoms.

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