Question #bfaae

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Question #bfaae

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Answer 1 · 2017-03-02T01:16:57

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

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

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