Subtopic: Water activity, sorption and desorption isotherms
Question Statement:
The moisture sorption isotherm of milk protein isolate follows the two-parameter BET equation: $\frac{a_w}{(1 - a_w) m} = \frac{1}{m_0 C} + \frac{C - 1}{m_0 C} a_w$. When $\frac{a_w}{(1 - a_w) m}$ is plotted against $a_w$, the linear regression yields an intercept of $I = 0.50\text{ kg dry matter/kg water}$ and a slope of $S = 9.50\text{ kg dry matter/kg water}$. The BET monolayer moisture content $m_0 = \frac{1}{S + I}$ in $\text{g water/100 g dry solid}$ is ________ (answer in integer).
Answer Key:
10
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Direct BET Parameter Substitution**
$\text{Intercept } I = \frac{1}{m_0 C} = 0.50$
$\text{Slope } S = \frac{C - 1}{m_0 C} = 9.50$
Sum of slope and intercept:
$S + I = \frac{C - 1}{m_0 C} + \frac{1}{m_0 C} = \frac{C}{m_0 C} = \frac{1}{m_0}$
$m_0 = \frac{1}{S + I} = \frac{1}{9.50 + 0.50} = \frac{1}{10.0} = 0.10\text{ kg water/kg dry solid}$
Converted to $\text{g/100 g dry solid}$:
$m_0 = 0.10 \times 100 = 10.0\text{ g/100 g dry solid}$
**Method 2: Determining C first**
$\frac{S}{I} = C - 1 = \frac{9.50}{0.50} = 19 \implies C = 20$
$m_0 = \frac{1}{C \cdot I} = \frac{1}{20 \times 0.50} = \frac{1}{10} = 0.10 = 10\%$