How would I find the molality of NaCl with the following information?
initial temperature= 98degrees celsius
final temp=100 degrees celsius.
Kbp(water)=0.512
To find the molality of NaCl, you need to use the freezing point depression equation. The equation is:
∆T = Kf * m * i
Where:
∆T is the change in temperature (in degrees Celsius),
Kf is the molal freezing point depression constant for the solvent (in this case, water),
m is the molality of the solute (NaCl in this case),
and i is the van't Hoff factor, which represents the number of particles into which the solute dissociates when it dissolves.
In this case, we can use the freezing point depression equation to find the molality of NaCl. However, we need to convert the given information to the appropriate units and determine the value of the van't Hoff factor.
First, let's calculate ∆T by subtracting the initial temperature from the final temperature:
∆T = 100°C - 98°C = 2°C
Next, we plug in the values for ∆T and Kf into the equation and solve for m:
2°C = Kf * m * i
Since the given information does not provide the van't Hoff factor, we can assume it to be 2 since NaCl dissociates into two ions in water.
2°C = Kf * m * 2
Now, rearrange the equation to solve for m:
m = 2°C / (Kf * 2)
m = 1°C / Kf
Lastly, replace Kf with the given value of the molal freezing point depression constant for water (Kbp = 0.512):
m = 1°C / 0.512
m ≈ 1.953 mol/kg
Therefore, the molality of NaCl is approximately 1.953 mol/kg.