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 = ?
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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