# What weight of solute is needed to prepare 350mL of 0.6M Na2CO2?

Aug 6, 2017

$18.9$ ${\text{g Na"_2"CO}}_{2}$

#### Explanation:

We're asked to find the mass of ${\text{Na"_2"CO}}_{2}$ needed to prepare $350$ $\text{mL}$ of a $0.6$ $M$ solution.

First, let's use the molarity equation:

$\underline{\overline{| \stackrel{\text{ ")(" ""molarity" = "mol solute"/"L solution"" }}{|}}}$

We're given:

• $\text{molarity} = 0.6$ $M$

• volume$= 350$ $\text{mL}$ $= 0.350$ $\text{L}$

Let's rearrange the equation to solve for the moles of solute:

"mol solute" = ("molarity")("L solution")

Plugging in known values:

${\text{mol Na"_2"CO"_2 = (0.6"mol"/(cancel("L")))(0.350cancel("L")) = color(red)(ul(0.21color(white)(l)"mol Na"_2"CO}}_{2}$

Now, we can use the molar mass of ${\text{Na"_2"CO}}_{2}$ ($89.989$ $\text{g/mol}$) to find the number of grams:

$\textcolor{red}{0.21} \cancel{\textcolor{red}{\text{mol Na"_2"CO"_2))((89.989color(white)(l)"g Na"_2"CO"_2)/(1cancel("mol Na"_2"CO"_2))) = color(blue)(ulbar(|stackrel(" ")(" "18.9color(white)(l)"g Na"_2"CO"_2" }} |}$

I'll leave it up to you (or your instructor) as to how many significant figures there should be.

Aug 6, 2017

Do you mean $N {a}_{2} C {O}_{2}$ or $N {a}_{2} C {O}_{3}$......??

#### Explanation:

$\text{Molarity"="Moles of solute (mol)"/"Volume of solution (L)}$

Thus $0.6 \cdot m o l \cdot {L}^{-} 1 = \text{moles of solute"/"350 mL}$....

and.................... $\text{mols of solute} = 0.6 \cdot m o l \cdot {L}^{-} 1 \times 350 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 = 0.21 \cdot m o l$

And this represents a mass of $0.21 \cdot m o l \times 105.99 \cdot g \cdot m o {l}^{-} 1$

$= 22.26 \cdot g$ with respect to $\text{sodium carbonate,}$ $N {a}_{2} C {O}_{3}$. There ain't no such beast as $N {a}_{2} C {O}_{2}$.