Question

You form a solution by dissolving 15.39 grams of sodium chloride (molar mass = 58.44 g/mol) into 163.7 grams of water (molar mass = 18.02 g/mol). The total volume of the solution is 180.1 mL.

Part #1:

What is the molarity of sodium chloride in this solution?

Part #2:

What is the mass percent of sodium chloride in this solution?

Part #3:

What is the mole fraction of sodium chloride in this solution?

Part #4:

What is the molality of sodium chloride in this solution?

Answer 1

Solution :-

lets first calculate moles of NaCl

Moles of NaCl = 15.39 g / 58.44 g per mol = 0.26335 mol

Volume of solution = 180.1 ml * 1 L / 1000 ml = 0.1801 L

Mass of solvent in kg = 163.7 g * 1 kgh / 1000 g = 0.1637 kg

Part #1:

What is the molarity of sodium chloride in this solution?

Molarity = moles / volume in liter

                 = 0.26335 mol / 0.1801 L

                 = 1.462 M

Part #2:

What is the mass percent of sodium chloride in this solution?

Mass % = (mass of NaCl/ total mass of solution )*100%

             = (15.39 g / (163.7 g + 15.39 g))*100%

             = 8.59 %

Part #3:

What is the mole fraction of sodium chloride in this solution?

Moles of water = 163.7 mol / 18.0148 g per mol = 9.087 mol

Mole fraction of NaCl = moles of NaCl / total moles

                                      = 0.26335 mol / (0.26335 mol + 9.087 mol )

                                      = 0.02816

Part #4:

What is the molality of sodium chloride in this solution?

Molality = moles / kg solvent

                = 0.26335 mol / 0.1637 kg

                = 1.609 m