Question

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

Answer 1

`18.9` `"g Na"_2"CO"_2`

Explanation:

We're asked to find the mass of `"Na"_2"CO"_2` needed to prepare `350` `"mL"` of a `0.6` `M` solution.

First, let's use the molarity equation:

`ulbar(|stackrel(" ")(" ""molarity" = "mol solute"/"L solution"" ")|)`

We're given:

• `"molarity" = 0.6` `M`

• volume`= 350` `"mL"` `= 0.350` `"L"`

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

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

Plugging in known values:

`"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 `"Na"_2"CO"_2` (`89.989` `"g/mol"`) to find the number of grams:

`color(red)(0.21)cancel(color(red)("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.

Answer 2

Do you mean `Na_2CO_2` or `Na_2CO_3`......??

Explanation:

`"Molarity"="Moles of solute (mol)"/"Volume of solution (L)"`

Thus `0.6*mol*L^-1="moles of solute"/"350 mL"`....

and.................... `"mols of solute"=0.6*mol*L^-1xx350*mLxx10^-3*L*mL^-1=0.21*mol`

And this represents a mass of `0.21*molxx105.99*g*mol^-1`

`=22.26*g` with respect to `"sodium carbonate,"` `Na_2CO_3`. There ain't no such beast as `Na_2CO_2`.