Answer to Question #249592 in General Chemistry for mas

Question

In the laboratory you dissolve 23.4 g of sodium fluoride in a volumetric flask and add water to a total volume of 375 mL.

What is the molarity of the solution?  ___________M.

What is the concentration of the sodium cation?  __________M.

What is the concentration of the fluoride anion? _______ M.

Answer 1

Amount of sodium fluoride, NaF added = 14.3 g

Molar mass of NaF = 41.988 g/mol

Hence, the number of moles of NaF taken is

n = mass molar mass 14.3 g 41.988 g/mol G= 0.3406 mol

Volume of the solution formed, V = 250 mL = 0.250 L

Now, molarity of a solution is defined as the number of moles of solute per litre of solution.

Hence, we can calculate the molarity of the solution as follows:

n 0.3406 mol molarity = 0 1.36 mol/L = 1.36 M .250 L

Hence, the molarity of the solution is 1.36 M.

Now, NaF dissociates as follows:

NaF (aq) + Natag) + Flag

Now, 1 mol of NaF dissociates to form 1 mol of Na+ and 1 mol of F- ions.

Hence, the concentration of Sodium cation and Fluoride anion will be same as the concentration of NaF solution.

Hence,

Concentration of sodium cation = 1.36 M

Concentration of Fluoride anion = 1.36 M.