Question

A brine solution for cheese making contains 228.5 g of sodium chloride (NaCl) in 500.0 mL of solution.

(a) Calculate the molarity of sodium chloride in the brine solution.

(b) How many liters of a stock solution of 18.28 M sodium chloride solution would be required to prepare 1.000 L of the brine solution? [Hint: Use the result from part (a).]

Answer 1

A brine solution for cheese making contains 228.5 g of sodium chloride (NaCl) in 500.0 mL of solution.

(a) Calculate the molarity of sodium chloride in the brine solution.

(b) How many liters of a stock solution of 18.28 M sodium chloride solution would be required to prepare 1.000 L of the brine solution? [Hint: Use the result from part (a).]

a) It is a concentration that states the number of moles of solute in exactly 1 L of solution.

Molariy(M) = moles of solute / Volume of Solution

1 mole of NaCl = 58.44 g of NaCl

Moles of NaCl = 228.5 * 1 mole of NaCl /58.44 g NaCl

                          = 3.909 M

But we have added 228.5 g of NaCl in 500 ml.

Therefore, molarity will be 3.909*2= 7.818 M

b)

Given

18.28 M NaCl solution

1.000 L brine solution with molarity of 7.818 M to be prepared.

Solution in L = 7.818 M * 1 L/18.28 M

                        = 0.4276 L