# Question d4f50

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 $\text{500 mL}$ of $L i t h i u m$ 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 $\text{0.175 moles}$ of $L i$ for every $\text{1 liter}$ of solution.

$\textcolor{w h i t e}{\text{aaaaaaaaaaaa}}$$\left(\text{0.175 moles" Li)/("1 Liter of solution}\right)$

The thing is, we don't have $\text{1 liter}$; we are told to prepare $\text{500 mL}$ . If you convert $\text{500 mL}$ to $L i t e r s$ first,

$\frac{500 \left(\cancel{m L}\right)}{1} \cdot \frac{\text{1 * "10^-3} L}{\cancel{m L}} = 0.500 L$

we get $\text{0.500 Liters} .$

$\textcolor{w h i t e}{\text{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}}$

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

$\textcolor{w h i t e}{\text{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}}$

We know that for every 1 mole of $L {i}_{3} P {O}_{4}$ will gives us $\text{3 moles of Li ions}$
$L {i}_{3} P {O}_{4} \to 3 L {i}^{+} + P O {4}^{-} 3$

This means $\text{0.087 moles of Li}$ will be released from one-third as many $L {i}_{3} P {O}_{4}$ molecules so $\text{0.029 moles}$ of $L {i}_{3} P {O}_{4}$

$\textcolor{w h i t e}{\text{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}}$

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

Molar mass of $L {i}_{3} P {O}_{4}$
$L i = 7 g \cdot 3 = 21 g$
$P = 31 g$
$O = 16 g \cdot 4 = 64 g$
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 $L {i}_{3} P {O}_{4}$