Question

An IV bag is labeled as 0.800% w/w sodium iodide in water solution with a density of 1.014g/mL. What is the concentration of sodium iodide in the solution expressed in molarity, molality, and mole fraction?

Molarity = ?

Molality = ?

Mole Fraction = ?

Answer #1

0.800% NaI in water solution:

100 g water contains 0.8 g NaI = 0.8/149.89 moles = 0.00534 moles (round off)

1000 g water contains 0.00534 moles x 1000 g /100 g = 0.0534 moles

**Molality = 0.0534 (m)**

**Mole Fraction =** 0.00534/ (0.00534 + 100/18) =
0.00534/ (0.00534 + 5.56) = 9.6 (round off).

Mass of the solution = (100 + 0.8) g

Density of the solution = 1.014 g/ml

Volume of the solution = (100 + 0.8) g/1.014 g/ml = 99.41 ml.

99.41 ml solution contains 0.00534 moles NaI

1000 ml solution contains 0.00534 moles x 1000 ml /99.41 ml = 0.054 moles

**Molarity = 0.054 (M)**

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