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

2 Answers
Aug 6, 2017

Answer:

#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.

Aug 6, 2017

Answer:

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#.