If 18 gram of glucose `(C_6H_(12)O_6)` is present in 1000 gram of an aqueous solution of glucose it is said to be-

Options

A. 1 molal
B. 1.1 molal
C. 0.5 molal
D. 0.1 molal

Answer

To solve the problem, we need to calculate the molality of the glucose solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- Mass of glucose (solute) = 18 grams
- Mass of the aqueous solution = 1000 grams

### Step 2: Calculate the mass of the solvent
To find the mass of the solvent, we subtract the mass of the solute from the total mass of the solution.
- Mass of solvent = Mass of solution - Mass of solute
- Mass of solvent = 1000 grams - 18 grams = 982 grams

### Step 3: Convert the mass of the solvent to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert grams to kilograms.
- Mass of solvent in kg = 982 grams / 1000 = 0.982 kg

### Step 4: Calculate the number of moles of glucose
To find the number of moles of glucose, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of glucose (C₆H₁₂O₆) can be calculated as follows:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 12 = 12 g/mol
- Oxygen (O): 16 g/mol × 6 = 96 g/mol
- Total molar mass of glucose = 72 + 12 + 96 = 180 g/mol

Now, calculate the number of moles:
- Number of moles of glucose = 18 g / 180 g/mol = 0.1 moles

### Step 5: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}} \]
Substituting the values we calculated:
- \( m = \frac{0.1 \text{ moles}}{0.982 \text{ kg}} \)

Calculating this gives:
- \( m \approx 0.1018 \)

### Step 6: Round to the appropriate significant figures
Since the question asks for a value that can be approximated, we can round this to:
- \( m \approx 0.1 \)

### Final Answer
The solution is said to be approximately **0.1 molal**.

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