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 of the following solutions at the same temperature will be isotonic, we need to compare their osmotic pressures. Isotonic solutions have the same osmotic pressure, which can be calculated using the formula:

\[ \pi = iCRT \]

Where:
- \(\pi\) = osmotic pressure
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(C\) = molarity (concentration in moles per liter)
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

### Step 1: Identify the Solutions
We have the following solutions to compare:
1. 3.42 g of cane sugar (C₁₂H₂₂O₁₁) in 1 liter of water
2. 0.18 g of glucose (C₆H₁₂O₆) in 1 liter of water

### Step 2: Calculate Molar Masses
- **Cane Sugar (C₁₂H₂₂O₁₁)**:
- Molar mass = \(12 \times 12 + 1 \times 22 + 16 \times 11 = 342 \, \text{g/mol}\)

- **Glucose (C₆H₁₂O₆)**:
- Molar mass = \(12 \times 6 + 1 \times 12 + 16 \times 6 = 180 \, \text{g/mol}\)

### Step 3: Calculate Molarity of Each Solution
- **For Cane Sugar**:
- Moles of cane sugar = \(\frac{3.42 \, \text{g}}{342 \, \text{g/mol}} = 0.01 \, \text{mol}\)
- Molarity (C) = \(\frac{0.01 \, \text{mol}}{1 \, \text{L}} = 0.01 \, \text{M}\)

- **For Glucose**:
- Moles of glucose = \(\frac{0.18 \, \text{g}}{180 \, \text{g/mol}} = 0.001 \, \text{mol}\)
- Molarity (C) = \(\frac{0.001 \, \text{mol}}{1 \, \text{L}} = 0.001 \, \text{M}\)

### Step 4: Compare Molarities
- Cane sugar: 0.01 M
- Glucose: 0.001 M

Since the molarities are not the same, these two solutions are not isotonic.

### Step 5: Check Other Options
Next, we check the other options provided in the question.

1. **3.42 g of cane sugar in 1 liter of water and 0.55 g of NaCl in 1 liter of water**:
- Molarity of NaCl:
- Moles of NaCl = \(\frac{0.55 \, \text{g}}{58.5 \, \text{g/mol}} = 0.0094 \, \text{mol}\)
- Molarity (C) = \(\frac{0.0094 \, \text{mol}}{1 \, \text{L}} = 0.0094 \, \text{M}\)
- For NaCl, \(i = 2\) (it dissociates into Na⁺ and Cl⁻):
- \(\pi = 2 \times 0.0094 \times RT = 0.0188RT\)
- Cane sugar: \(\pi = 0.01RT\)
- Not isotonic.

2. **3.42 g of cane sugar in 1 liter of water and 1.17 g of NaCl in 1 liter of water**:
- Molarity of NaCl:
- Moles of NaCl = \(\frac{1.17 \, \text{g}}{58.5 \, \text{g/mol}} = 0.02 \, \text{mol}\)
- Molarity (C) = \(\frac{0.02 \, \text{mol}}{1 \, \text{L}} = 0.02 \, \text{M}\)
- For NaCl, \(i = 2\):
- \(\pi = 2 \times 0.02 \times RT = 0.04RT\)
- Cane sugar: \(\pi = 0.01RT\)
- Not isotonic.

### Conclusion
The only isotonic solutions are:
- **3.42 g of cane sugar in 1 liter of water and 0.18 g of glucose in 1 liter of water**.

### Final Answer
The correct option is the second one: **3.42 g of cane sugar in 1 liter of water and 0.18 g of glucose in 1 liter of water are isotonic solutions.**