To solve the problem step by step, we will follow the process of calculating the elevation in boiling point, determining the number of moles of glucose, and finally calculating the number of molecules of glucose in the solution.
### Step 1: Calculate the elevation in boiling point (ΔTb)
The boiling point of the solution is given as 100.07°C. The normal boiling point of water is 100°C.
To find the elevation in boiling point:
\[
\Delta T_b = T_{\text{solution}} - T_{\text{pure water}} = 100.07°C - 100°C = 0.07°C
\]
### Step 2: Convert the elevation in boiling point to Kelvin
Since the elevation in boiling point is the same in Celsius and Kelvin for small differences, we can keep it as:
\[
\Delta T_b = 0.07 \, \text{K}
\]
### Step 3: Use the formula for boiling point elevation
The formula for boiling point elevation is given by:
\[
\Delta T_b = K_b \cdot m
\]
Where:
- \( K_b \) is the molal elevation constant (0.5 K kg mol⁻¹)
- \( m \) is the molality of the solution
### Step 4: Calculate the molality (m)
We can rearrange the formula to find molality:
\[
m = \frac{\Delta T_b}{K_b}
\]
Substituting the values:
\[
m = \frac{0.07 \, \text{K}}{0.5 \, \text{K kg mol}^{-1}} = 0.14 \, \text{mol/kg}
\]
### Step 5: Calculate the number of moles of glucose
Molality is defined as the number of moles of solute per kilogram of solvent. Given that we have 100 g of water, we convert this to kg:
\[
\text{mass of water} = 100 \, \text{g} = 0.1 \, \text{kg}
\]
Now we can calculate the number of moles of glucose:
\[
\text{Number of moles of glucose} = m \cdot \text{mass of solvent (kg)} = 0.14 \, \text{mol/kg} \cdot 0.1 \, \text{kg} = 0.014 \, \text{mol}
\]
### Step 6: Calculate the number of molecules of glucose
To find the number of molecules, we use Avogadro's number, which is \( 6.02 \times 10^{23} \, \text{molecules/mol} \):
\[
\text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} = 0.014 \, \text{mol} \times 6.02 \times 10^{23} \, \text{molecules/mol}
\]
Calculating this gives:
\[
\text{Number of molecules} \approx 8.42 \times 10^{21} \, \text{molecules}
\]
### Final Answer
The number of molecules of glucose in the solution containing 100 g of water is approximately \( 8.42 \times 10^{21} \) molecules.
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