Give step-by-step solution with explanation and final answer:
(give me answer without using bullet points)Plot the boiling point elevation AT, as a function of the weight fraction of a non-volatile s

Question

Give step-by-step solution with explanation and final answer:
(give me answer without using bullet points)Plot the boiling point elevation AT, as a function of the weight fraction of a non-volatile solute
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VIPSolutions · Accepted Answer

Let's address **the first problem step-by-step**:

---

### **Problem Statement**
*Plot the boiling point elevation ΔT_b as a function of the weight fraction of a non-volatile solute in an aqueous solution, accounting for the van't Hoff factor i. ΔT_b = i·K_b·m. Establish the concentration at which the boiling point of the solution exceeds that of the pure solvent by exactly 2.5 K.*

---

### **Step 1: Write the Boiling Point Elevation Equation**

The boiling point elevation is given by:
$$
\Delta T_b = i K_b m
$$
Where:
- $ \Delta T_b $ = boiling point elevation (K)
- $ i $ = van't Hoff factor (unitless)
- $ K_b $ = ebullioscopic constant of the solvent (for water, $ K_b \approx .512 $ K·kg/mol)
- $ m $ = molality of the solution (mol solute/kg solvent)

---

### **Step 2: Express Molality in Terms of Weight Fraction**

Let:
- $ w_2 $ = weight fraction of solute
- $ w_1 $ = weight fraction of solvent ($ w_1 + w_2 = 1 $)
- $ M_2 $ = molar mass of solute (g/mol)
- $ M_1 $ = molar mass of solvent (for water, $ M_1 = 18.015 $ g/mol)

The molality $ m $ is:
$$
m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{\dfrac{w_2}{M_2}}{\dfrac{w_1}{M_1}} \cdot \frac{M_1}{w_1} \cdot \frac{1}{1}
$$
But more simply, with 1 kg of *total* solution,
$$
m = \frac{w_2 / M_2}{w_1}
$$
where $ w_1 $ is in kg.

But since $ w_2 $ is *weight fraction* (unitless), for 1 kg total solution:
$$
\text{Mass of solute} = w_2 \quad\textrm{kg}
$$
$$
\text{Mass of solvent} = w_1 = 1 - w_2 \quad\textrm{kg}
$$
$$
m = \frac{w_2 / M_2}{1 - w_2}
$$

---

### **Step 3: Substitute m into ΔT_b Expression**

$$
\Delta T_b = i K_b \frac{w_2}{M_2 (1 - w_2)}
$$

---

### **Step 4: Set ΔT_b = 2.5 K and Solve for w_2**

$$
2.5 = i K_b \frac{w_2}{M_2 (1 - w_2)}
$$

$$
2.5 M_2 (1 - w_2) = i K_b w_2
$$

$$
2.5 M_2 - 2.5 M_2 w_2 = i K_b w_2
$$

$$
2.5 M_2 = i K_b w_2 + 2.5 M_2 w_2
$$

$$
2.5 M_2 = w_2 (i K_b + 2.5 M_2)
$$

$$
w_2 = \frac{2.5 M_2}{i K_b + 2.5 M_2}
$$

---

### **Step 5: Final Answers**

#### **(a) Plot equation (as a function of weight fraction):**
$$
\boxed{
\Delta T_b = i K_b \frac{w_2}{M_2 (1 - w_2)}
}
$$

#### **(b) Weight fraction for ΔT_b = 2.5 K:**
$$
\boxed{
w_2 = \frac{2.5 M_2}{i K_b + 2.5 M_2}
}
$$
where $ i $ = van't Hoff factor, $ K_b $ = ebullioscopic constant, $ M_2 $ = molar mass of solute.

---

#### **Explanation**
You start with the boiling point elevation law, express molality in terms of the weight fraction, substitute, and then solve algebraically for the desired weight fraction that yields a 2.5 K elevation. Plug in values for $ i $, $ K_b $, and $ M_2 $ as needed for a specific solute.

If you need the plot or numerical example for a particular solute, provide its $ M_2 $ and $ i $.

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