# Question bfaae

Mar 2, 2017

a) $1.4 \cdot {10}^{- 3}$ M F
b) $1.14 \cdot {10}^{13}$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 $1.4 \cdot {10}^{- 3}$g. Therefore, its molarity is $1.4 \cdot {10}^{- 3}$ M in this case.

To reach a $1.4 \cdot {10}^{- 3}$ M concentration of F in 1.4×10^8 L requires
1.4×10^8# L * $1.4 \cdot {10}^{- 3}$ mol/L = $1.96 \cdot {10}^{11}$ 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
$1.96 \cdot {10}^{11}$ mol * 58g/mol = $113.7 \cdot {10}^{11}$g , or $1.14 \cdot {10}^{13}$g KF