Question

5.00g of a certain manufacturer's baking powder was placed in 10mL of water and analyzed in the laboratory through atomic absorption spectroscopy. The analysis found the baking powder solution contained 5.20 wt% NaCl with a density of 1.09g/mL. From the manufacturer's label, 1 serving of baking powder equals of 0.5g of the baking powder. How many milligrams of sodium are present in one serving of this particular baking powder?

Answer 1

Ans. Given-

            5.00 g of baking powder is dissolved in 10.0 mL water

            [NaCl] in resultant solution is 5.20 wt%, with density of 1.09 g mL^-1

# Step 1: Quantify mass of Na in the solution:

            Mass of solution = Vol. x Density = 10.0 mL x 1.09 g mL^-1 = 10.9 g

            Mass of NaCl in the solution = [NaCl] wt% x Mass of solution

                                                            = 5.20 % of 10.9 g = 0.5668 g

            Moles of NaCl in the soln. = Mass / Molar mass

                                                            = 0.5668 g / 58.44 g mol^-1 = 0.0097 mol

# Now, each NaCl molecule consists of 1 Na-atom. So, the moles of Na in the solution must be equal to that of NaCl.

            So, Moles of Na in the solution = 0.0097 mol

And,

            Mass of Na in the soln. = Moles x Molar mass

                                                = 0.0097 mol x 22.99 g mol^-1 = 0.223003 g = 223 mg

# Since the solution is prepared from 5.00 g of baking powder, the total Na content of 5.00 g sample must be equal to that of the solution.

            Hence, Na content in 5.00 g baking powder = 223 mg

# Step 2: Quantifying Na per serving

From #step 1, we have-       Na content in 5.00 g baking powder = 223 mg

Given-             Baking powder serving size = 0.50 g baking powder per serving

Now,

Na content in 1 serving = [Na] in baking powder x Serving size

                                    = (223 mg Na / 5.0 g baking powder) x 0.50 g baking powder

                                    = 22.3 mg