What is the molality of a solution which contains 36 g of glucose `(C_(4)H_(12)O_(5))` in 250 g of water ?

Options

A. `0.8m`
B. 0.16m
C. 0.4m
D. .11m

Explanation

To find the molality of the solution containing 36 g of glucose (C₆H₁₂O₆) in 250 g of water, we can follow these steps:

### Step 1: Convert the mass of water to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of water from grams to kilograms.

\[ \text{Mass of water} = 250 \, \text{g} = \frac{250}{1000} \, \text{kg} = 0.25 \, \text{kg} \]

### Step 2: Calculate the molar mass of glucose
The molar mass of glucose (C₆H₁₂O₆) can be calculated by adding the atomic masses of its constituent elements:
- 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:
\[ \text{Molar mass of glucose} = 72.06 + 12.096 + 96.00 = 180.156 \, \text{g/mol} \approx 180 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of glucose
Now, we can calculate the number of moles of glucose using the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of glucose} = \frac{36 \, \text{g}}{180 \, \text{g/mol}} = 0.2 \, \text{moles} \]

### Step 4: Calculate the molality of the solution
Now we can calculate the molality (m) using the formula:
\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Substituting the values:
\[ \text{Molality} = \frac{0.2 \, \text{moles}}{0.25 \, \text{kg}} = 0.8 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is **0.8 mol/kg**.

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