What weight of solute is needed to prepare 350mL of 0.6M Na2CO2?
2 Answers
Answer:
Explanation:
We're asked to find the mass of
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
#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:
Do you mean
Explanation:
Thus
and....................
And this represents a mass of