## Chemistry: Molecular Approach (4th Edition)

$\underline{3.11\times {{10}^{5}}\text{ lb}}$
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 $\underline{3.11\times {{10}^{5}}\text{ lb}}$.