Answer
$8.0\times10^{4} kg\ NaF$
Work Step by Step
Data provided:
Population 150,000 people
Water consumption 660 L/person.day
Water density $1.0\ g/cm^3 = 1.0\ kg/L$
Fluoride concentration in water $1\times10^{-6} kg\ F^-/kg\ water$
Fluoride mass percentage in sodium fluoride $45.0\%=0.450\ kg\ F^-/kg\ NaF$
Yearly water consumption in kg $150,000\ people \times 660\ L/person.day \times 365\ days/year\times1.0\ kg/L=3.6\times10^{10} kg$
Yearly fluoride consumption $3.6\times10^{10} kg \times 1\times10^{-6} kg\ F^-/kg\ water = 3.6\times10^{4} kg\ F^-$
Yearly NaF consumption $3.6\times10^{4} kg \div0.450\ kg\ F^-/kg\ NaF = 8.0\times10^{4} kg\ NaF$
