NaCl dissociates completely in water. Therefor the total number of particles in a 1.0 m NaCl solution is no longer 6.02 X 10^23. Since a colligative property is dependent only on the total number of solute particles, what is the freezing point for 1.0 m NaCl solution?
nevermind. i figured it out
To determine the freezing point of a 1.0 m NaCl solution, we need to understand the concept of colligative properties and how they are influenced by the number of solute particles.
Colligative properties are physical properties of a solution that depend on the concentration of solute particles, rather than their type. These properties include freezing point depression, boiling point elevation, vapor pressure reduction, and osmotic pressure.
In the case of a 1.0 m NaCl solution, it means that we have 1.0 mole of NaCl dissolved in 1 liter of water. However, NaCl dissolves completely into its constituent ions, Na+ and Cl-. So, in reality, we have 1.0 mole of Na+ ions and 1.0 mole of Cl- ions in the solution.
Since each NaCl formula unit dissociates into two ions, the total number of solute particles in the solution becomes 2.0 moles.
Now, we can use the colligative property equation for freezing point depression, which states:
ΔTf = Kf * m
Where:
ΔTf is the change in freezing point,
Kf is the cryoscopic constant (specific for each solvent),
m is the molality of the solution (moles of solute per kilogram of solvent).
The cryoscopic constant for water is approximately 1.86 °C/m.
In this case, since the concentration is given as 1.0 m, we have 1.0 mole of solute (2.0 moles of particles) dissolved in 1 kilogram of solvent (water).
Substituting the values into the equation:
ΔTf = 1.86 °C/m * 2.0 mol/kg = 3.72 °C
Therefore, the freezing point of the 1.0 m NaCl solution is depressed by 3.72 °C compared to pure water.