Answer
the weight of the helium is $667.829 \mathrm{lb}$
Work Step by Step
Calculate the mass of the helium as follows:
$$
m=\rho \times V
$$
Here, the Volume of helium is denoted by $V$.
Substitute $68000 \mathrm{ft}^{3}$ for $V$ and $3.05 \times 10^{-4}$ slugs $/ \mathrm{ft}^{3}$ for $\rho$
$$
\begin{aligned}
m &=3.05 \times 10^{-4} \text { slugs } / \mathrm{ft}^{3} \times 68000 \mathrm{ft}^{3} \\
&=20.74 \text { slugs }
\end{aligned}
$$
Calculate the weight of the helium as follows:
$$
W=m g
$$
Here, the Acceleration due to gravity is $g$.
Now, substitute $20.74$ slugs for $m$
$$
\begin{aligned}
W &=20.74 \text { slugs } \times 32.2 \\
&=667.829 \mathrm{lb}
\end{aligned}
$$
Therefore, the weight of the helium is $667.829 \mathrm{lb}$
