Question

JEE MAINS 2010
Which of the following has the largest

number of atoms ?

a. 1g Au (s) b. 1g Na (s)

c. 1g Li (s) d. 1g Cl2 (g)​

Answer 1

Answer:

third option is the correct one

Answer 2

$\Large\boxed{\sf{\quad(c)\quad\!1g\ Li_{(s)}\quad}}$

Solution 1:-

• GMM of Au = 197 g

Hence no. of moles in 1 g Au $\sf{=\dfrac{1}{197}\ mol.}$

So the no. of atoms in 1 g Au $\sf{=\dfrac{1}{197}\times N_A.}$

• GMM of Na = 23 g.

Hence no. of moles in 1 g Na $\sf{=\dfrac{1}{23}\ mol.}$

So the no. of atoms in 1 g Na $\sf{=\dfrac{1}{23}\times N_A.}$

• GMM of Li = 7 g.

Hence no. of moles in 1 g Li $\sf{=\dfrac{1}{7}\ mol.}$

So the no. of atoms in 1 g Li $\sf{=\dfrac{1}{7}\times N_A.}$

• GMM of Cl₂ = 71 g.

Hence no. of moles in 1 g Cl₂ $\sf{=\dfrac{1}{71}\ mol.}$

So the no. of atoms in 1 g Cl₂ $\sf{=\dfrac{1}{71}\times 2N_A=\dfrac{2}{71}\times N_A}$

[2 is the atomicity of Cl₂]

Now the compounds in decreasing order of their no. of moles is,

$\longrightarrow\sf{\dfrac{1}{7}\times N_A\ \textgreater\ \dfrac{1}{23}\times N_A\ \textgreater\ \dfrac{2}{71}\times N_A\ \textgreater\ \dfrac{1}{197}\times N_A}$

That is,

$\longrightarrow\sf{Li\ \textgreater\ Na\ \textgreater\ Cl\ \textgreater\ Au}$

Hence 1 g Li contains the largest no. of atoms.

Solution 2:-

The decreasing order of the no. of atoms in given compounds is equivalent to the increasing order of their gram atomic masses (GAM's), at same mass of the compounds.

• GAM of Au = 197 g

• GAM of Na = 23 g

• GAM of Li = 7 g.

• GAM of Cl = 35.5 g

So the increasing order of GAM's is,

$\longrightarrow\sf{7\ \textless\ 23\ \textless\ 35.5\ \textless\ 197}$

That is,

$\longrightarrow\sf{Li\ \textless\ Na\ \textless\ Cl\ \textless\ Au}$

Hence Li has the least GAM and hence 1 g Li contains the largest no. of atoms.