The expected boiling point of a solution prepared by dissolving 4.46 g of sodium iodide (nai) in 51.8 g of water (h2o) is 100.084 °C
To determine the expected boiling point of a solution prepared by dissolving 4.46 g of sodium iodide in 51.8 g of water, you can use the following equation:
ΔTb = Kb * molality
Where ΔTb is the change in boiling point, Kb is the molal boiling point constant for water (0.52 °C/m), and molality is the concentration of the solute in the solution.
First, you need to calculate the molality of the solution. To do this, you can use the following equation:
molality = moles of solute / kilograms of solvent
Plugging in the values, you get:
molality = (4.46 g / 214.00 g/mol) / (51.8 g / 1000 g/kg)
Solving, you get:
molality = 0.0161 m
Next, plug this value into the equation above to calculate the change in boiling point:
ΔTb = 0.52 °C/m * 0.0161 m
Solving, you get:
ΔTb = 0.084 °C
Finally, add this value to the boiling point of pure water (100 °C) to determine the expected boiling point of the solution:
Boiling point = 100 °C + 0.084 °C
Solving, you get:
Boiling point = 100.084 °C
Therefore, the expected boiling point of a solution prepared by dissolving 4.46 g of sodium iodide in 51.8 g of water is approximately 100.084 °C.
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