Question

Question #bfaae

Answer 1

a) `1.4 * 10^(-3)` M F
b) `1.14 * 10^13`g KF

Explanation:

Molarity is the number of moles per liter of solution. The molecular (atomic) weight of fluorine is 19g/mol, or 19000 mg/mol. 1.4mg F is `1.4 * 10^(-3)`g. Therefore, its molarity is `1.4 * 10^(-3)` M in this case.

To reach a `1.4 * 10^(-3)` M concentration of F in `1.4×10^8` L requires
`1.4×10^8` L * `1.4 * 10^(-3)` mol/L = `1.96 * 10^11` mol F
Delivered as KF, we would need the same number of moles of KF. Adding “K” to the “F” weight previously used, we have a molecular weight of 58g/mol. The amount of grams of KF that will need to be added is then
`1.96 * 10^11` mol * 58g/mol = `113.7 * 10^11`g , or `1.14 * 10^13`g KF