Calculating ion molarity using solute mass

Question

A chemist prepares a solution of sodium chloride (NaCl) by measuring out 70.mg of NaCl into a 300.mL  volumetric flask and filling to the mark with distilled water.

Calculate the molarity of Cl− anions in the chemist's solution.

Be sure your answer is rounded to 2 significant digits.

Answer 1

1. Molarity of sodium chloride. ( Mol wt is 58.44)

58.44 grams in 1L is 1 M

58.44 mg in 1 mL is 1M

17532 mg in 300 mL is 1M

Rx

70 mg in 300 mL is 0.00399 M [ 70 mg x 1 M / 17532 mg = 0.00399 M]

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Next determination of molairty of Cl^- ion

Molarity of Cl^- = Molaity of NaCl x ion/solute ratio

Ion/solute ratio = mol of Cl^- / mol of NaCl = 1 mol of Cl^- / 1 mol of NaCl

Molarity of Cl^- = 0.00399 M NaCl x 1 mol Cl^- / 1 mol NaCl = 0.00399 M or 0.00399 mol Cl^- / L

= 0.0040 mol Cl^- /L

Answer 2

Hi Oleg!

To find the molarity of the Cl^- ions in solution, we must first recall the equation for molarity.

Molarity= ___moles of solute_

liters of solution

Now, we have to convert the 70mg of NaCl into moles and the 300mL of water into liters so that these values can be plugged in to our equation.

70mg NaCl x 1 gram NaCl x 1 mole NaCl = 0.00120 mol NaCl

1 1000mg NaCl 58.44g NaCl

300mL sol x 1 Liter sol = 0.300 L sol

1 1000mL sol

Lastly, we can plug our value for moles and liters into the molarity equation and solve for the molarity of the solution. In this particular solution, the molarity of the chloride anions is equal to the molarity of the solution because when we look at the chemical equation for the breakdown of NaCl we notice that the NaCl and Cl^- are in a one to one ratio with each other. So our answer should be:

0.00120 mol Cl^- = 0.004 M Cl^-

0.300 L solution

I hope this helps!