Question

An IV bag is labeled as 0.800% w/w sodium chloride in water solution with a density of 1.036g/mL. What is the concentration of sodium chloride in the solution expressed in molarity, molality, and mole fraction?

Answer 1

Answer:

[NaCl] = 0.14 M

[NaCl] = 0.14 m

Mole fraction of NaCl → 2.48×10⁻³

Explanation:

This is a problem of concentration.

0.8% w/w means that in 100 g of solution, we have 0.8 g of solute, in this case NaCl.

Let's determine the volume with the density.

Solution density = Solution mass / Solution volume

1.036 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.036 g/mL → 96.5 mL

Let's calculate molarity (mol/L)

We convert the mass to moles (mass / molar mass)

0.8 g / 58.45 g/mol = 0.0137 moles

We must convert the volume of solution to L.

Molarity is mol/L, moles of solute in L of solution.

96.5 mL . 1L/1000 mL = 0.0965 L

0.0137 mol / 0.0965 L = 0.14 M

Let's determine molality and mole fraction.

Molality are moles of solute in 1 kg of solvent (mol/kg)

Mass of solution = Mass of solute + Mass of solvent

100 g = 0.8 g + Mass of solvent

100 g - 0.8 g = Mass of solvent → 99.2 g

Then, we must convert the mass of solvent to kg

99.2 g . 1kg / 1000 g = 0.0992 kg

Molality: 0.0137 mol / 0.0992 kg  → 0.14 m

Mole fraction → moles of solute / moles of solute + moles of solvent

Let's find out the moles of solvent ( mass / molar mass)

99.2g / 18 g/mol = 5.511 mol

Total moles = 5.511 + 0.0137 → 5.5247 moles

Mole fraction = 0.0137 / 5.5247 → 2.48×10⁻³