The molality of the solution containing 65g of glucose dissolved in 2kg of water is ?

Explanation

To find the molality of the solution containing 65g of glucose dissolved in 2kg of water, we can follow these steps:

### Step 1: Calculate the number of moles of glucose.
To find the number of moles of glucose, we use the formula:

\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]

The molar mass of glucose (C₆H₁₂O₆) is calculated as follows:
- Carbon (C): 12.01 g/mol × 6 = 72.06 g/mol
- Hydrogen (H): 1.008 g/mol × 12 = 12.096 g/mol
- Oxygen (O): 16.00 g/mol × 6 = 96.00 g/mol

Adding these together gives:

\[ \text{Molar mass of glucose} = 72.06 + 12.096 + 96.00 = 180.156 \approx 180 \text{ g/mol} \]

Now, substituting the values into the moles formula:

\[ \text{Moles of glucose} = \frac{65 \text{ g}}{180 \text{ g/mol}} \approx 0.361 \text{ moles} \]

### Step 2: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent:

\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]

Given that the mass of the solvent (water) is 2 kg, we can substitute the values:

\[ \text{Molality} = \frac{0.361 \text{ moles}}{2 \text{ kg}} \approx 0.1805 \text{ mol/kg} \]

### Final Answer:
The molality of the solution is approximately **0.18 mol/kg**.

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