Which of the following solutions at the same temperature will be isotonic?

Options

A. 3.42 g of cane sugar in one litre water and 0.18 g of glucose in one litre water.
B. 3.42 g of cane sugar in one litre water and 0.18 g of glucose in 0.1 litre water.
C. 3.42 g of cane sugar in one litre water and 0.585 g of NaCl in one litre water.
D. 3.42 g of cane sugar in one litre water and 1.17 g of NaCl in one litre water.

Answer

To determine which solutions are isotonic at the same temperature, we need to analyze the osmotic pressure of each solution. Isotonic solutions have the same osmotic pressure, which means they have the same effective concentration of solute particles in solution.

### Step-by-Step Solution:

1. **Understand Osmotic Pressure**:
The osmotic pressure (π) is given by the formula:
\[ \pi = C_{\text{effective}} \cdot R \cdot T \]
where \(C_{\text{effective}} = C \cdot i\), \(C\) is the molarity, and \(i\) is the Van't Hoff factor (number of particles the solute dissociates into).

2. **Identify Given Solutions**:
We will analyze the provided options, which include different solutes and their respective weights in specific volumes of solution.

3. **Calculate Molarity for Each Solution**:
- For **Cane Sugar (C₁₂H₂₂O₁₁)**:
- Weight = 3.42 g, Molecular Weight = 342 g/mol
- Moles of Cane Sugar = \(\frac{3.42 \text{ g}}{342 \text{ g/mol}} = 0.01 \text{ mol}\)
- Volume = 1 L, Molarity = \(\frac{0.01 \text{ mol}}{1 \text{ L}} = 0.01 \text{ M}\)
- Van't Hoff factor \(i = 1\) (does not dissociate).

- For **Glucose (C₆H₁₂O₆)**:
- Weight = 0.18 g, Molecular Weight = 180 g/mol
- Moles of Glucose = \(\frac{0.18 \text{ g}}{180 \text{ g/mol}} = 0.001 \text{ mol}\)
- Volume = 1 L, Molarity = \(\frac{0.001 \text{ mol}}{1 \text{ L}} = 0.001 \text{ M}\)
- Van't Hoff factor \(i = 1\).

- For **NaCl**:
- Weight = 0.585 g, Molecular Weight = 58.5 g/mol
- Moles of NaCl = \(\frac{0.585 \text{ g}}{58.5 \text{ g/mol}} = 0.01 \text{ mol}\)
- Volume = 1 L, Molarity = \(\frac{0.01 \text{ mol}}{1 \text{ L}} = 0.01 \text{ M}\)
- Van't Hoff factor \(i = 2\) (dissociates into Na⁺ and Cl⁻).

4. **Calculate Effective Concentration**:
- For Cane Sugar: \(C_{\text{effective}} = 0.01 \cdot 1 = 0.01\)
- For Glucose: \(C_{\text{effective}} = 0.001 \cdot 1 = 0.001\)
- For NaCl: \(C_{\text{effective}} = 0.01 \cdot 2 = 0.02\)

5. **Compare Osmotic Pressures**:
- Cane Sugar: \(\pi = 0.01 \cdot R \cdot T\)
- Glucose: \(\pi = 0.001 \cdot R \cdot T\)
- NaCl: \(\pi = 0.02 \cdot R \cdot T\)

6. **Determine Isotonic Solutions**:
- The only solution with the same osmotic pressure would be the one with the same effective concentration. In this case, Cane Sugar and Glucose do not match in effective concentration, and NaCl has a higher effective concentration due to dissociation.

### Conclusion:
The only isotonic solution among the options provided is **Option 2**, where the molarity of both solutes is the same.