How many moles of NaCl are contained in 467 mL of a .244M solution of NaCl?

1 Answer
Dec 12, 2015

Answer:

#"0.144 moles"#

Explanation:

As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride, #"NaCl"#, divided by liters of solution.

#color(blue)("molarity" = "moles of solute"/"liters of solution")#

Simply put, dissolving one mole of a solute in one liter of solution will produce a #"1.00-M"# solution.

In your case, the volume of the solution is said to be equal to #"467 mL"# and the molarity of the solution to #"0.244 M"#. Since molarity tells you number of moles per liter, you can say that you will get #"0.244 moles"# of sodium chloride in one liter of this solution.

Since you have a little less than half of a liter, i.e. #"467 mL"#, you can expect to have fewer than half of #"0.244 moles"#.

More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!

#color(blue)(c = n/V implies n = c * V)#

#n = 0.244 "moles"/color(red)(cancel(color(black)("L"))) * 467 * 10^(-3) color(red)(cancel(color(black)("L"))) = "0.1139 moles NaCl"#

Rounded to three sig figs, the answer will be

#n_(NaCl) = color(green)("0.114 moles")#

SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a #"0.244 M"# sodium chloride solution.

Instead, you would say that you have a solution that is #"0.244 M"# in sodium cations, #"Na"^(+)#, and #"0.244 M"# in chloride anions, #"Cl"^(-)#.