The molality of the solution containing 65g of glucose dissolved in 1kg of water is ?

Options

A. 0.36m
B. 0.56m
C. 0.45m
D. 015m

Explanation

To find the molality of the solution containing 65g of glucose dissolved in 1kg 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:

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

Given:
- Mass of glucose (solute) = 65 g
- Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol

Now, substituting the values:

\[ \text{Number of moles of glucose} = \frac{65 \, \text{g}}{180 \, \text{g/mol}} \approx 0.3611 \, \text{moles} \]

### Step 2: Identify the mass of the solvent
The mass of the solvent (water) is given as:

\[ \text{Mass of solvent} = 1 \, \text{kg} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:

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

Substituting the values we calculated:

\[ \text{Molality} = \frac{0.3611 \, \text{moles}}{1 \, \text{kg}} = 0.3611 \, \text{mol/kg} \]

### Step 4: Round the result
For simplicity, we can round the result to two decimal places:

\[ \text{Molality} \approx 0.36 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is approximately **0.36 mol/kg**.

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