If a mole were defined to be 3 × 10²⁴ instead of Avogadro's number (N_A), then what would be the atomic mass of Argon?
Atomic mass of Argon is 40 (on conventional sale).

Atomic mass of argon is the mass of argon atoms present in its mole and its value is given as 40. Number of argon atoms present in one mole of argon atoms is $6.023\times {10}^{23}.$ This indicates that the mass of $6.023\times {10}^{23} \mathrm{argon} \mathrm{atoms}$ is 40.

$\begin{array}{ccc}\mathrm{Number} \mathrm{of} \mathrm{argon} \mathrm{atoms} \mathrm{present} \mathrm{in} \mathrm{one} \mathrm{mole}& =& 6.023\times {10}^{23} \mathrm{atoms}\\ \mathrm{Mass} \mathrm{of} \mathrm{one} \mathrm{mole} \mathrm{of} \mathrm{argon} \mathrm{atom}& =& 40\\ \mathrm{Mass} \mathrm{of} 6.023\times {10}^{23} \mathrm{argon} \mathrm{atoms}& =& 40\end{array}$

 If a mole is defined as $3\times {10}^{24} \mathrm{atoms}$ instead of $6.023\times {10}^{23} \mathrm{atoms},$ then mass of $3\times {10}^{24} \mathrm{argon} \mathrm{atoms}$ is determined as:

 $\begin{array}{ccc}\mathrm{Mass} \mathrm{present} \mathrm{in} 3\times {10}^{24} \mathrm{argon} \mathrm{atoms}& =& 3\times {10}^{24} \mathrm{argon} \mathrm{atoms}\times \frac{40}{6.023\times {10}^{23} \mathrm{argon} \mathrm{atoms}}\\ & =& 199\end{array}$