The freezing point depression of a .100 m solution of NaCl solution is .34C. Calculate the percent dissociation of NaCl.( Kf for water=1.86)
Please help
If you assume that freezing point depression is dependent on just dissociation, then
DeltaTf= Kf*molality*(percentdissociation)*number particlesperformula
Here, the particlesperformula is 2 (Na + Cl), so solve for percent dissociation, and you will get a decimal percent. multiply itby 100 to get percent. In my head, get about 90 percent here.
To calculate the percent dissociation of NaCl, we need to first understand the concept of freezing point depression and its relationship to the dissociation of solutes.
Freezing point depression occurs when a solute is added to a solvent, resulting in a lower freezing point than the pure solvent. The extent of freezing point depression depends on both the concentration of the solute and its dissociation in the solvent.
In this case, we are given that the freezing point depression (∆Tf) of a 0.100 m solution of NaCl is 0.34°C, and the cryoscopic constant (Kf) for water is 1.86°C/m.
First, let's calculate the molality (m) of the NaCl solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Since we are given the concentration of the solution (0.100 m), we can assume 1 kg of solvent. Therefore:
m = 0.100 m
Next, we can use the formula for freezing point depression:
∆Tf = Kf * m
Let's plug in the values:
0.34°C = 1.86°C/m * 0.100 m
Now, we can solve for the moles of NaCl that dissociate in the solution. Since NaCl dissociates into one Na+ ion and one Cl- ion, the moles of NaCl dissociated will be twice the moles of the solute.
Let's denote the moles of NaCl as x:
moles of NaCl dissociated = 2 * x
Now, let's use the equation for freezing point depression to relate the moles of NaCl dissociated and the decrease in freezing point:
∆Tf = Kf * m = 2 * x
We know that ∆Tf = 0.34°C, and we can substitute the value of m:
0.34°C = 1.86°C/m * 0.100 m = 2x
Simplifying the equation:
0.34°C = 0.186°C * x
Now, we can solve for x, which represents the moles of NaCl dissociated:
x = 0.34°C / 0.186°C = 1.828 moles
Since we assumed that 1 kg of solvent was used, and the molar mass of NaCl is 58.44 g/mol, we can find the mass of NaCl:
mass of NaCl = 1.828 moles * 58.44 g/mol = 106.799 g
Finally, we can calculate the percent dissociation of NaCl:
percent dissociation = (moles dissociated / total moles) * 100
In this case:
moles dissociated = 1.828 moles
total moles = 1 mol (since 1 NaCl molecule dissociates into 1 Na+ ion and 1 Cl- ion)
percent dissociation = (1.828 moles / 1 mol) * 100 = 182.8%
Therefore, the percent dissociation of NaCl in the 0.100 m solution is 182.8%.