How many moles of koh are required to prepare 0.80 l of 6.4 m koh? 0.13 mol koh 5.1 mol koh 7.2 mol koh 8.0 mol koh?

Question

Ryder

Chemistry

3 March, 08:59

How many moles of koh are required to prepare 0.80 l of 6.4 m koh? 0.13 mol koh 5.1 mol koh 7.2 mol koh 8.0 mol koh?

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Answers (1)

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Rolando Bernard · Answer 1

Answer is: 5,1 mol KOH.

V (KOH) = 0,8l = 0,8 dm³.

c (KOH) = 6,4 mol/dm³.

V - volume of solution.

c - concentration of solution.

c = n:V

n - amount of substance.

n (KOH) = c·V = 6,4 mol/dm³ · 0,8 dm³ = 5,12 mol.

Potassium hydroxide (KOH) can dissolve in water at room temperature.