= Solution
Convert the <molar mass> to $0.0288\ \mathrm{kg\,mol^{-1}}$ before substitution:
$$
\boxed{\frac{dT}{dz}=-\frac{(0.0288)(9.81)}{29}
=-9.742\times10^{-3}\ \mathrm{K\,m^{-1}}
\simeq-9.74\ \mathrm{K\,km^{-1}}.}
$$
The often-used environmental value near $-6.5\ \mathrm{K\,km^{-1}}$ is less steep because rising moist air can condense <water vapor>, releasing <latent heat> and offsetting some expansion cooling. A saturated parcel therefore follows a smaller-magnitude <moist adiabatic lapse rate>, which varies with <temperature>, <pressure> and <water> content. Real environmental profiles also reflect radiation, mixing and weather systems; $6.5\ \mathrm{K\,km^{-1}}$ is a representative profile value, not a universal parcel law.
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