How many moles of sodium carbonate is present in a 325 mL solution with a molar concentration of 0.620 M?

a) 1.91 mol

B) 2.02 mol

C) 0.524 mol

D) 0.202 mol

1 Answer
Shantelle
May 12, 2018

#D) " "0.202 " mol"#

Explanation:

#M#, or molarity, is #"moles solute"/"liters solution"#.

We have #325# #"mL"# solution, and we know that is equivalent to #0.325# #L#, as #325# #"mL"# #xx "1 L"/"1000 mL" = "0.325 L"#.

Therefore, to find the moles of sodium carbonate present, we do:
#"0.620 mol solute"/"L solution " xx "0.325 L solution"#

#= "0.2015 mols"# sodium carbonate

However, we can only have 3 significant figures (sig figs), meaning we have round to the thousandths place in this case, rounding the answer to #"0.202 mols"# sodium carbonate.

Therefore, the answer is #D)#.

Hope this helps!

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