# Step-by-Step Solution Let's break down the problem and solve it step by step using **graphical method** as requested--- ## **Given Data** |₀ (number/cc.s) | Nucleation Temp (°C | Solubility (wt%)|------------------|----------------------|------------------| | 30000 | 25 | 10 | | 15000 | 35 | 25 | | 700 | 45 | 40 | | 300 | 55 | 60 | - **Initial concentration of solution**: 70 wt% - **Temperature to be cooled to**: 15°C - **Solubility of MgSO₄ at 15°C**: 6 wt% --- ## **Step 1: Understanding Supersaturation** Supersaturation, \( S \), can be defined as: \[ S = \frac{C - C^*}{C^*} \] where: - \( C \): Actual concentration - \( C^* \): Solubility at given temperature However, for graphical methods we often use just the values of concentration and solubility. --- ## **Step 2: Find Supersaturation for Each Table Entry** For each nucleation temperature, calculate the supersaturation using the given solubility and a hypothetical concentration equal to the initial (but we'll use the solubility for plotting): But for our case, the actual relevant supersaturation at 15°C is: - \( C = 70 \) wt% (initial) - \( C^* = 6 \) wt% (solubility at 15°C) So, supersaturation at 15°C: \[ S_{15^\circ C} = \frac{70 - 6}{6} = \frac{64}{6} \approx 10.67 \] --- ## **Step 3: Prepare Data for Graph** For graphical method, we plot **nucleation rate (B₀) vs supersaturation**. For each table point: \[ S = \frac{C - C^*}{C^*} \] But the table seems to assume that the solution is saturated at each temperature (i.e., concentration = solubility at that T). So, let's extract the data points: - At 25°C: C = 10, C* = 10, S = - At 35°C: C = 25, C* = 25, S = - At 45°C: C = 40, C* = 40, S = - At 55°C: C = 60, C* = 60, S = But for our case, the solution is cooled to **15°C** with a much higher concentration (70 wt%). So, we need to **extrapolate** using the pattern. --- ## **Step 4: Graphical Method (Extrapolation)** Since we cannot draw here, let's describe how you would do it: 1. **On graph paper:** - X-axis: Supersaturation (C - C*) or C - Y-axis: Nucleation Rate B₀ 2. **Plot the given points:** - (10, 30000) - (25, 15000) - (40, 700) - (60, 300) 3. **Extrapolate to C = 70 (for 15°C)** --- ## **Step 5: Interpolate/Extrapolate B₀ at C = 70** We see that as concentration increases, B₀ decreases. Let's estimate the trend (it looks roughly linear on a semi-log plot). Let's fit the data: \[ \begin{align*} \text{C (wt%)} & : 10, 25, 40, 60 \\ \text{B}_ & : 30000, 15000, 700, 300 \\ \end{align*} \] Let’s use logarithmic regression (since B₀ drops rapidly): \[ \log_{10} B_ = a - bC \] Using two points to estimate \(a\) and \(b\): - At C=10, B₀=30000: \(\log_{10} 30000 = 4.477\) - At C=60, B₀=300: \(\log_{10} 300 = 3.477\) So, \[ 4.477 = a - b \times 10 \\ 3.477 = a - b \times 60 \] Subtract: \[ 4.477 - 3.477 = b(60 - 10) \\ 1. = 50b \\ b = .02 \] Now solve for \(a\): \[ 4.477 = a - .02 \times 10 \implies a = 4.677 \] So, the equation: \[ \log_{10} B_ = 4.677 - .02 C \] Now substitute \(C = 70\): \[ \log_{10} B_ = 4.677 - .02 \times 70 = 4.677 - 1.4 = 3.277 \] \[ B_ = 10^{3.277} \approx 190 \text{ number/cc.s} \] --- ## **Step 6: Final Answer** **The nucleation rate of MgSO₄ in the solution cooled to 15°C (with 70 wt% concentration) is approximately:** \[ \boxed{190 \ \text{number/cc.s}} \] --- ### **Summary of Steps** 1. Find the supersaturation at 15°C. 2. Prepare a graph of nucleation rate vs. concentration or supersaturation. 3. Fit a trend (logarithmic regression). 4. Extrapolate to the desired concentration (70 wt%). 5. Calculate B₀. --- **If you have graph paper, you can plot the actual data points and read off the value at 70 wt% for a more visual solution!**