Question

Which one of the following will have the largest number of atoms? $($Molar mass of $Au=197g/mol)$

• A. $1\text{g of}Li\text{(s)}$
• B. $1\text{g}Au\text{(s)}$
• C. $1\text{g of}C{l}_2\text{(g)}$
• D. $1\text{g}Na\text{(s)}$

Answer 1

The correct option is A $1\text{g of}Li\text{(s)}$

The number of atoms can be calculated as
$\text{(a)}1\text{g}Au=\frac{1}{197}\times 6.022\times {10}^{23}=\frac{6.022}{197}\times {10}^{23}$
$\text{(b)}1\text{g}Na=\frac{6.022}{23}\times {10}^{23}$
$\text{(c)}1\text{g}Li=\frac{6.022}{7}\times {10}^{23}$
$\text{(d)}1\text{g}C{l}_2=\frac{2\times 6.022\times {10}^{23}}{71}=\frac{6.022}{35.5}\times {10}^{23}$
$1\text{g}$ of $Li$ contains largest number of atoms.