Question #d4f50

1 Answer
Mar 30, 2017

Let's look at what we have.

Explanation:

We have 2 givens in the question stem and we are solving for mass.

We need to prepare #"500 mL"# of #Lithium# ion solution with a concentration of #0.175 M.# Well, what does that mean exactly?

You have a given volume with a given number of moles. That basically translates to saying we have #"0.175 moles"# of #Li# for every #"1 liter"# of solution.

#color (white)"aaaaaaaaaaaa"##("0.175 moles" Li)/("1 Liter of solution")#

The thing is, we don't have #"1 liter"#; we are told to prepare #"500 mL"# . If you convert #"500 mL"# to #Liters# first,

#(500(cancel(mL)))/(1) * ("1 * "10^-3" L)/(cancel(mL)) = 0.500 L#

we get #"0.500 Liters".#

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Using our conversion factor we can solve for moles.

#(0.500 cancel("Liters of solution"))/1 * ("0.175 moles" Li)/(1 cancel("Liter of solution")) = "0.087 moles" Li#

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We know that for every 1 mole of #Li_3PO_4# will gives us #"3 moles of Li ions"#
#Li_3PO_4 ->3 Li^+ + PO4^-3#

This means #"0.087 moles of Li"# will be released from one-third as many #Li_3PO_4# molecules so #"0.029 moles"# of #Li_3PO_4#

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Now find the molar mass and convert from moles to mass to figure out the unknown.

Molar mass of #Li_3PO_4#
#Li = 7 g * 3 = 21g#
#P = 31 g#
#O = 16 g * 4 = 64g#
Total: 116 g

#(0.029 cancel("moles" Li_3PO_4))/1 * (116 grams)/(1 cancel("mole" Li_3PO_4)) = "3.364 grams of" Li_3PO_4#

Answer: 3.364 grams of #Li_3PO_4#