To determine the expected change in the freezing point of water in a solution of [tex]\( 62.5 \)[/tex] g of barium nitrate ([tex]\(Ba(NO_3)_2\)[/tex]) in 1 kg of water, we need to follow several steps. Here's a detailed step-by-step calculation:
### Step 1: Calculate the number of moles of [tex]\(Ba(NO_3)_2\)[/tex]
The given mass of barium nitrate is [tex]\(62.5 \)[/tex] g and the molar mass of [tex]\(Ba(NO_3)_2\)[/tex] is [tex]\(261 \)[/tex] g/mol. The number of moles can be determined using the formula:
[tex]\[ \text{moles of } Ba(NO_3)_2 = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{moles of } Ba(NO_3)_2 = \frac{62.5 \text{ g}}{261 \text{ g/mol}} \][/tex]
[tex]\[ \text{moles of } Ba(NO_3)_2 \approx 0.239 \text{ moles} \][/tex]
### Step 2: Determining the Van't Hoff factor
Barium nitrate ([tex]\(Ba(NO_3)_2\)[/tex]) dissociates completely in water to form barium ions ([tex]\(Ba^{2+}\)[/tex]) and nitrate ions ([tex]\(NO_3^-\)[/tex]). The dissociation can be represented as:
[tex]\[ Ba(NO_3)_2 \rightarrow Ba^{2+} + 2 NO_3^- \][/tex]
This results in 3 ions (1 [tex]\(Ba^{2+}\)[/tex] ion and 2 [tex]\(NO_3^-\)[/tex] ions).
So, the Van't Hoff factor [tex]\(i\)[/tex] for [tex]\(Ba(NO_3)_2\)[/tex] is:
[tex]\[ i = 3 \][/tex]
### Step 3: Calculate the freezing point depression
The formula for freezing point depression ([tex]\(\Delta T_f\)[/tex]) is:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
where:
- [tex]\( i \)[/tex] is the Van't Hoff factor
- [tex]\( K_f \)[/tex] is the freezing point depression constant for water ([tex]\(1.86 \ ^\circ C \cdot kg/mol\)[/tex])
- [tex]\( m \)[/tex] is the molality of the solution
The molality [tex]\(m\)[/tex] of the solution is calculated as:
[tex]\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \][/tex]
Here, the solvent is water and its mass is 1 kg:
[tex]\[ m = \frac{0.239 \text{ moles}}{1 \text{ kg}} \][/tex]
[tex]\[ m = 0.239 \text{ mol/kg} \][/tex]
Now, substitute these values into the freezing point depression formula:
[tex]\[ \Delta T_f = 3 \cdot 1.86 \ ^\circ C \cdot kg/mol \cdot 0.239 \text{ mol/kg} \][/tex]
[tex]\[ \Delta T_f \approx 1.336 \ ^\circ C \][/tex]
### Conclusion
The freezing point of water decreases by approximately [tex]\(1.336 \ ^\circ C\)[/tex], which means the change in the freezing point is [tex]\(-1.336 \ ^\circ C\)[/tex].
Among the given options, the closest correct answer is:
a. [tex]\(-1.34 ^\circ C\)[/tex]
