Freezing Point Depression and Boiling Point Elevation Wksht - Quia - Free Printable

Here are the step-by-step solutions for each problem on your worksheet.

1. Freezing Point of Sucrose Solution

Goal: Find the change in freezing point ($\Delta T_f$).

* Step 1: Find the molar mass of sucrose ($C_{12}H_{22}O_{11}$).
* Carbon (C): $12 \times 12.01 = 144.12$
* Hydrogen (H): $22 \times 1.01 = 22.22$
* Oxygen (O): $11 \times 16.00 = 176.00$
* Total Molar Mass $\approx 342.34 \text{ g/mol}$

* Step 2: Calculate moles of sucrose.
* $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.68 \text{ g}}{342.34 \text{ g/mol}} \approx 0.01075 \text{ mol}$

* Step 3: Calculate Molality ($m$).
* $\text{Molality} = \frac{\text{moles solute}}{\text{kg solvent}} = \frac{0.01075 \text{ mol}}{2.50 \text{ kg}} \approx 0.00430 \text{ m}$

* Step 4: Calculate $\Delta T_f$.
* Formula: $\Delta T_f = i \cdot K_f \cdot m$
* Sucrose is a covalent compound (non-electrolyte), so it doesn't split apart. The van 't Hoff factor ($i$) is 1.
* $\Delta T_f = 1 \cdot (-1.86^\circ\text{C/m}) \cdot 0.00430 \text{ m}$
* $\Delta T_f \approx -0.0080^\circ\text{C}$

2. Boiling Point of Sodium Fluoride Solution

Goal: Find the change in boiling point ($\Delta T_b$).

* Step 1: Determine the van 't Hoff factor ($i$).
* Sodium fluoride (NaF) splits into two ions: $Na^+$ and $F^-$.
* So, $i = 2$.

* Step 2: Calculate Molality ($m$).
* We already have moles ($5.76 \text{ mol}$) and mass of water ($3.62 \text{ kg}$).
* $m = \frac{5.76 \text{ mol}}{3.62 \text{ kg}} \approx 1.591 \text{ m}$

* Step 3: Calculate $\Delta T_b$.
* Formula: $\Delta T_b = i \cdot K_b \cdot m$
* $\Delta T_b = 2 \cdot (0.51^\circ\text{C/m}) \cdot 1.591 \text{ m}$
* $\Delta T_b \approx 1.62^\circ\text{C}$

3. Freezing Point of Potassium Iodide Solution

Goal: Find the change in freezing point ($\Delta T_f$).

* Step 1: Find the molar mass of KI.
* Potassium (K): $39.10$
* Iodine (I): $126.90$
* Total Molar Mass $= 166.00 \text{ g/mol}$

* Step 2: Calculate moles of KI.
* $\text{Moles} = \frac{30.0 \text{ g}}{166.00 \text{ g/mol}} \approx 0.1807 \text{ mol}$

* Step 3: Calculate Molality ($m$).
* $m = \frac{0.1807 \text{ mol}}{1.75 \text{ kg}} \approx 0.1033 \text{ m}$

* Step 4: Determine $i$ and calculate $\Delta T_f$.
* KI splits into $K^+$ and $I^-$, so $i = 2$.
* $\Delta T_f = 2 \cdot (-1.86^\circ\text{C/m}) \cdot 0.1033 \text{ m}$
* $\Delta T_f \approx -0.384^\circ\text{C}$

4. Boiling Point of Salt Water

Goal: Find the change in boiling point ($\Delta T_b$).

* Step 1: Find the molar mass of Salt (NaCl).
* Sodium (Na): $22.99$
* Chlorine (Cl): $35.45$
* Total Molar Mass $= 58.44 \text{ g/mol}$

* Step 2: Calculate moles of NaCl.
* $\text{Moles} = \frac{30.0 \text{ g}}{58.44 \text{ g/mol}} \approx 0.5133 \text{ mol}$

* Step 3: Calculate Molality ($m$).
* $m = \frac{0.5133 \text{ mol}}{3.75 \text{ kg}} \approx 0.1369 \text{ m}$

* Step 4: Determine $i$ and calculate $\Delta T_b$.
* NaCl splits into $Na^+$ and $Cl^-$, so $i = 2$.
* $\Delta T_b = 2 \cdot (0.51^\circ\text{C/m}) \cdot 0.1369 \text{ m}$
* $\Delta T_b \approx 0.14^\circ\text{C}$

5. Finding $K_f$ from Depression

Goal: Find the constant $K_f$.

* Step 1: Identify variables.
* Molality ($m$) = $3.60 \text{ m}$
* Change in Temp ($\Delta T_f$) = $-0.851^\circ\text{C}$
* Assume non-electrolyte unless stated otherwise, so $i = 1$.

* Step 2: Rearrange formula.
* $\Delta T_f = i \cdot K_f \cdot m$
* $K_f = \frac{\Delta T_f}{i \cdot m}$

* Step 3: Solve.
* $K_f = \frac{-0.851}{1 \cdot 3.60}$
* $K_f \approx -0.236^\circ\text{C/m}$

6. Finding $K_b$ from Elevation

Goal: Find the constant $K_b$.

* Step 1: Identify variables.
* Molality ($m$) = $5.70 \text{ m}$
* Change in Temp ($\Delta T_b$) = $1.62^\circ\text{C}$
* Assume non-electrolyte, so $i = 1$.

* Step 2: Rearrange formula.
* $\Delta T_b = i \cdot K_b \cdot m$
* $K_b = \frac{\Delta T_b}{i \cdot m}$

* Step 3: Solve.
* $K_b = \frac{1.62}{1 \cdot 5.70}$
* $K_b \approx 0.284^\circ\text{C/m}$

7. Freezing Point of Camphor Solution

Goal: Find the change in freezing point ($\Delta T_f$).

* Step 1: Find the molar mass of dimethyl ether ($C_2H_6O$).
* Carbon (C): $2 \times 12.01 = 24.02$
* Hydrogen (H): $6 \times 1.01 = 6.06$
* Oxygen (O): $1 \times 16.00 = 16.00$
* Total Molar Mass $= 46.08 \text{ g/mol}$

* Step 2: Calculate moles of solute.
* $\text{Moles} = \frac{10.0 \text{ g}}{46.08 \text{ g/mol}} \approx 0.2170 \text{ mol}$

* Step 3: Calculate Molality ($m$).
* $m = \frac{0.2170 \text{ mol}}{3.00 \text{ kg}} \approx 0.07233 \text{ m}$

* Step 4: Calculate $\Delta T_f$.
* Dimethyl ether is covalent, so $i = 1$.
* Given $K_f = 5.95^\circ\text{C/m}$. Note: Freezing point depression is a negative change.
* $\Delta T_f = 1 \cdot (-5.95^\circ\text{C/m}) \cdot 0.07233 \text{ m}$
* $\Delta T_f \approx -0.430^\circ\text{C}$

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Final Answer:
1. -0.0080°C
2. 1.62°C
3. -0.384°C
4. 0.14°C
5. -0.236°C/m
6. 0.284°C/m
7. -0.430°C