Question

Sea water contains around $65\times {10}^{-3}$ g of bromide ions in $1$ dm${}^3$. If all the bromide ions are converted into ${Br}_2$, then how many dm${}^3$ of sea water are needed to produce 1kg of bromide ${Br}_2$?

• A. 3076.9${dm}^3$
• B. 30769 $\times {10}^{-3}{dm}^3$
• C. 30769$\times {10}^3{dm}^3$
• D. 30769${dm}^3$
• E. None of the above

Answer 1

The correct option is D 30769${dm}^3$

${Br}^{-}⟶\frac{1}{2}{Br}^2+e^{-}$
$65\times {10}^{-3}g{Br}^{-}=32.5\times {10}^{-3}g{Br}_2$ is in ${1dm}^3$
$\therefore 1000g{Br}_2$ is in $=\frac{1000}{32.5\times {10}^{-3}}=30769{dm}^3$