Question

What would be the final volume (in mL) of the aqueous potassium fluoride which was prepared by dissolving 8.0 g of potassium fluoride in an appropriate volume of water to give a 0.89 M aqueous solution?

Answer 1

0.15 L

Explanation:

Step 1. Calculate the moles of KF.

Molar mass = 58.10 g/mol

Moles of KF = 8.0 × 1/58.10

Moles of KF = 0.138 mol KF

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Step 2. Calculate the volume of KF

c = n/V Multiply both sides by V

V = 0.138 × 1/0.89

V = 0.15 L