About how much should the boiling point of pure water differ from .5m 1m and 1.5m aqueous solutions of NaCl and same molal solutions of CaCl2 Trying to check experimental data from van't hoff factor lab
You can calculate this from
delta T = i*Kb*m
i = 2 for NaCl and 3 for CaCl2.
To determine the approximate difference in boiling points between pure water and the given solutions, you need to consider the concept of boiling point elevation caused by the presence of solutes. This can be calculated using the van 't Hoff factor (i) and the molality (m) of the solutions.
The van 't Hoff factor (i) represents the number of particles into which a solute dissociates in a solution. For NaCl, i is approximately 2, meaning it dissociates into two ions (Na+ and Cl-) when dissolved in water. For CaCl2, i is approximately 3, as it dissociates into three ions (Ca2+ and 2Cl-) in water.
The equation used to calculate boiling point elevation is:
ΔTb = i * Kb * m
Where:
- ΔTb represents the change in boiling point
- Kb is the molal boiling point constant (for water, it is approximately 0.52 °C/m)
- m is the molality of the solution (moles of solute per kilograms of solvent)
Let's use this information to calculate the approximate difference in boiling points:
For a 0.5m NaCl solution:
- ΔTb = 2 * 0.52 °C/m * 0.5 m
- ΔTb ≈ 0.52 °C
For a 1m NaCl solution:
- ΔTb = 2 * 0.52 °C/m * 1 m
- ΔTb ≈ 1.04 °C
For a 1.5m NaCl solution:
- ΔTb = 2 * 0.52 °C/m * 1.5 m
- ΔTb ≈ 1.56 °C
For a 0.5m CaCl2 solution:
- ΔTb = 3 * 0.52 °C/m * 0.5 m
- ΔTb ≈ 0.78 °C
For a 1m CaCl2 solution:
- ΔTb = 3 * 0.52 °C/m * 1 m
- ΔTb ≈ 1.56 °C
For a 1.5m CaCl2 solution:
- ΔTb = 3 * 0.52 °C/m * 1.5 m
- ΔTb ≈ 2.34 °C
Keep in mind that these values are approximate and based on the assumptions of ideal behavior of the solutions. Experimental values may have some variations due to factors like impurities or non-ideal behavior. It's always best to compare these calculations with your experimental data to check the accuracy.