0.1 M solution of a solute has a water potential of

Options

A. -2.3 bars
B. Zero
C. 2.3 bars
D. 22.4 bars

Answer

To find the water potential of a 0.1 M solution of a solute, we can use the formula for water potential (Ψ):

\[ \Psi = -iCRT \]

Where:
- \( \Psi \) = Water potential
- \( i \) = Ionization constant (number of particles the solute dissociates into)
- \( C \) = Molar concentration of the solution (in molarity)
- \( R \) = Pressure constant (0.0831 liter bar per mole kelvin)
- \( T \) = Temperature in Kelvin (for standard conditions, we can use 273 K)

### Step-by-Step Solution:

1. **Identify the Ionization Constant (i)**:
- For a non-electrolyte solute, the ionization constant \( i \) is typically 1. If the solute dissociates into more than one particle, you would use that number. Here, we assume \( i = 1 \).

2. **Determine the Molar Concentration (C)**:
- The problem states that the concentration of the solution is 0.1 M. So, \( C = 0.1 \).

3. **Use the Pressure Constant (R)**:
- The value of \( R \) is given as 0.0831 liter bar per mole kelvin.

4. **Set the Temperature (T)**:
- We will use standard temperature, which is 273 K.

5. **Plug the values into the water potential formula**:
\[ \Psi = -iCRT \]
\[ \Psi = -1 \times 0.1 \, \text{M} \times 0.0831 \, \text{liter bar per mole kelvin} \times 273 \, \text{K} \]

6. **Calculate the value**:
- First, calculate \( 0.1 \times 0.0831 \times 273 \):
\[ 0.1 \times 0.0831 = 0.00831 \]
\[ 0.00831 \times 273 = 2.27 \, \text{(approximately)} \]
- Therefore, \( \Psi = -2.27 \, \text{bars} \).

7. **Round the value**:
- Rounding to one decimal place gives us approximately \( -2.3 \, \text{bars} \).

### Final Answer:
The water potential of a 0.1 M solution of the solute is approximately **-2.3 bars**.

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