The vapour pressure of pure water at `26^@C` is 25.5 torr. . The vapour pressure of a solution which contains 20.0 glucose, `(C_6H_12O_6)` , in 100 g water (in torr) is ?

Answer

To solve the problem of finding the vapor pressure of a solution containing 20 g of glucose in 100 g of water, we can follow these steps:

### Step 1: Determine the vapor pressure of pure water
The vapor pressure of pure water at 26°C is given as \( P_0 = 25.5 \, \text{torr} \).

### Step 2: Calculate the number of moles of glucose
To find the number of moles of glucose, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of glucose (\( C_6H_{12}O_6 \)) is approximately 180 g/mol. Thus, the number of moles of glucose is:
\[ \text{Number of moles of glucose} = \frac{20 \, \text{g}}{180 \, \text{g/mol}} = 0.1111 \, \text{mol} \]

### Step 3: Calculate the number of moles of water
The molar mass of water (\( H_2O \)) is approximately 18 g/mol. Therefore, the number of moles of water is:
\[ \text{Number of moles of water} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.5556 \, \text{mol} \]

### Step 4: Calculate the mole fraction of the solute (glucose)
The mole fraction of glucose (\( X_{solute} \)) can be calculated using the formula:
\[ X_{solute} = \frac{\text{Number of moles of glucose}}{\text{Number of moles of glucose} + \text{Number of moles of water}} \]
Substituting the values:
\[ X_{solute} = \frac{0.1111}{0.1111 + 5.5556} = \frac{0.1111}{5.6667} \approx 0.0196 \]

### Step 5: Use Raoult's Law to find the vapor pressure of the solution
According to Raoult's Law:
\[ P = P_0 \times (1 - X_{solute}) \]
Where \( P \) is the vapor pressure of the solution. Thus:
\[ P = 25.5 \, \text{torr} \times (1 - 0.0196) = 25.5 \, \text{torr} \times 0.9804 \approx 25.0 \, \text{torr} \]

### Final Answer
The vapor pressure of the solution is approximately **25.0 torr**.