Question

A 2.900×10−2M solution of NaCl in water is at 20.0 ºC. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.2 mL. The density of water at 20.0 °C is 0.9982 g/mL. What is the mole fraction of salt in this solution?

A) 5.23 x 10⁻⁴
B) 99.95 x 10⁻²
C) 5.23 x 10⁻²
D) 99.95 x 10⁻⁴

Answer 1

The mole fraction of salt in the solution is 0.345 or 34.5%.

Step-by-step explanation:

To calculate the mole fraction of salt in the solution, we first need to find the moles of NaCl present in the solution and the moles of water. The moles of NaCl can be calculated by multiplying the molarity of the solution by the volume of the solution in liters:

Moles of NaCl = Molarity × Volume = (2.900×10−2 M) × (1.000 L) = 0.029 moles

The moles of water can be calculated by converting the volume of water from milliliters to liters and then dividing by the molar mass of water:

Moles of water = Volume of water (in liters) ÷ Molar mass of water = (999.2 mL × 0.001 L/mL) ÷ (18.015 g/mol) = 0.055 moles

The mole fraction of salt in the solution is then calculated by dividing the moles of NaCl by the total moles of solute:

Mole fraction of salt = Moles of NaCl ÷ (Moles of NaCl + Moles of water) = 0.029 moles ÷ (0.029 moles + 0.055 moles) = 0.345 or 34.5%