Chemistry and Chemical Reactivity (9th Edition)

$\rho_{Ar}=1.78\ g/L$
The problem states that the air density is the sum of the products of densities of its constituents by their volume fraction, so from the data given: Air (dry, $CO_2$ free): $\rho= 1.29327 \ g/L$, $20.96\%\ O_2,\ 78.11\%\ N_2,\ 0.930\%\ Ar$ $O_2$:$\rho= 1.42952\ g/L$ $N_2$:$\rho= 1.25092\ g/L$ So: $1.29327=20.96/100\times1.42952+78.11/100\times1.25092+0.930/100\times \rho_{Ar}$ $\rho_{Ar}=1.78\ g/L$