Answer
a) $49 \ kg$
b) $2.5 \times 10^3 \ kg$
Work Step by Step
We know the result from the last problem:
$m_g = \frac{\rho_g M}{\rho_a-\rho_g}$
Thus, we find:
a) $m = \frac{(280)(.18)}{1.22-.18}=49 \ kg$
b) $m = \frac{(280)(.9\rho_a)}{\rho_a-.9\rho_a}=2.5 \times 10^3 \ kg$
