atoms - Effects of Changing Avogadro's Constant

The Avogadro project suggests that we redefine the Avogadro constant to be equal to our best known estimate, $N_\mathrm{A} = 6.02214179 \times 10^{23}$, and redefine the kilogram based on the Avogadro constant.

What happens if we define the Avogadro constant to be exactly $6 \times 10^{23}$ and reassign atomic weights correspondingly?

Would any unforeseen complications arise from here?

Example:

Under current definitions, $N_\mathrm{A}$ atoms of carbon-12 weigh 12 g. If we redefine the constant to have the value $N_\mathrm{A}^\prime = 6.0\times 10^{23}$, then $N_\mathrm{A}^\prime$ atoms of carbon-12 would weigh $\pu{11.955879 g}$, which we would have to assign as the new atomic mass of carbon-12.

Answer

As of May 20th 2019, the Avogadro constant will be set to $6.02214076 \times 10^{23}\ \rm{mol}^{–1}$, i.e. the Avogadro constant times one mole will be an integer, but not the one the OP asked about. In preparing for the switch of definition, the BIPM looked carefully at two ways of determining the constant, and then set it to the best estimate from experimental values, truncated to nine significant figures.

This way, there will be minimal practical changes of units even though the definitions have changed. Also changed is what is defined vs. what is measured. For example, the molar mass of 12-C is now based on measurement, and could change (certainly in terms of how many significant figures are known) as measurement techniques improve.

Setting the Avogadro constant to a dimensionless number as the OP asked about would make lots of relationships between quantities and constant dimensionally incorrect, such as $$ R = k_\mathrm{B} N_\mathrm{A}$$

with $k_\mathrm{B}$ the Boltzmann constant, R the universal gas constant and $N_\mathrm{A}$ the Avogadro constant.

Would any unforeseen complications arise from here?

Setting Avogadro's constant to $6 \times 10^{23}\ \rm{mol}^{–1}$ (I added the proper units) would have lots of complications, but they would not be unforeseen, as @Zhe discusses in the comments.