Answer
\[\underline{3.11\times {{10}^{5}}\text{ lb}}\]
Work Step by Step
Volume of water in the pool is 185 cubic yards.
\[1\text{ y}{{\text{d}}^{3}}=764,555\text{ c}{{\text{m}}^{3}}\]
Thus, volume of 185 cubic yards in cubic centimeter will be written as follows:
\[\begin{align}
& V=\left( 764,555\times 185 \right)\text{ c}{{\text{m}}^{3}} \\
& =141,442,675\text{ c}{{\text{m}}^{3}}
\end{align}\]
The relation between density \[\left( d \right)\] and mass \[\left( m \right)\] is as follows:
\[m=\left( V \right)\left( d \right)\]
Here, V is volume.
Density of water is \[1\ g\text{/c}{{\text{m}}^{3}}\]. Thus, mass of water is calculated as follows:
\[\begin{align}
& m=\left( V \right)\left( d \right) \\
& =\left( 141,442,675\text{ c}{{\text{m}}^{3}} \right)\left( 1\text{ g/c}{{\text{m}}^{3}} \right) \\
& =141,442,675\text{ g}
\end{align}\]
Now,
\[\text{1 g}=0.0022\text{ lb}\]
Thus, mass of water in pounds is written as follows:
\[\begin{align}
& m=\left( 141,442,675\times 0.0022 \right)\text{ lb} \\
& =311,173.89\text{ lb} \\
& =\text{3}\text{.11}\times \text{1}{{\text{0}}^{5}}\text{ lb}
\end{align}\]
The mass of water in pounds is \[\underline{3.11\times {{10}^{5}}\text{ lb}}\].
