Question

Which one of the following will have the largest number of atoms?

$i$ $1gAu\left(s\right)$

$ii$ $1gNa\left(s\right)$

$iii$ $1gLi\left(s\right)$

$iv$ $1gC{l}_2\left(s\right)$

Answer 1

We know,

Number of moles of compound $=\frac{\text{Given mass of compound}}{\text{Molar mass of compound}}$

$\left(i\right)1g\text{of Au(s)}=\frac{1}{197}\text{mol of Au(s)}$
$=\frac{1}{197}\times 6.022\times {10}^{23}\text{atoms of Au(s)}$
$=3.06\times {10}^{21}\text{atoms of Au(s)}$

$\left(ii\right)1gofNa\left(s\right)=\frac{1}{23}mol\text{of}Na\left(s\right)$
$\frac{1}{23}\times 6.022\times {10}^{23}\text{atoms of}Na\left(s\right)$
$26.2\times {10}^{21}\text{atoms of}Na\left(s\right)$

$\left(iii\right)1gLi\left(s\right)=\frac{1}{7}mole\text{of}Li\left(s\right)$
$=6.022\times {10}^{23}\times \frac{1}{7}\text{atoms of}Li\left(s\right)$
$86\times {10}^{21}\text{atoms of}Li\left(s\right)$

$\left(iv\right)1gC{l}_2\left(g\right)=\frac{1}{71}molofC{l}_2\left(g\right)$
$\frac{1}{71}\times 6.022\times {10}^{23}\times 2\text{atoms of}Cl$
$16.96\times {10}^{21}\text{atoms of}Cl$
Hence, $1g\text{of}Li\left(s\right)$ has the largest number of atoms.

Final Answer: Among the given substances, $1g\text{of}Li\left(s\right)$ will have the
largest number of atoms.