A solution of glucose `(C_(6)H_(12)O_(6))` in water is labelled as 10% by weight . What would be the molality of the solution ?

Explanation

To find the molality of a 10% by weight glucose solution in water, we can follow these steps:

### Step 1: Understand the meaning of 10% by weight
A 10% by weight solution means that there are 10 grams of glucose in every 100 grams of the solution.

### Step 2: Determine the weight of the solvent (water)
Since the total weight of the solution is 100 grams and the weight of glucose is 10 grams, the weight of the solvent (water) can be calculated as:
\[ \text{Weight of solvent} = \text{Total weight of solution} - \text{Weight of solute (glucose)} = 100 \, \text{g} - 10 \, \text{g} = 90 \, \text{g} \]

### Step 3: Calculate the number of moles of glucose
To find the number of moles of glucose, we use the formula:
\[ \text{Number of moles} = \frac{\text{Weight of solute}}{\text{Molecular weight of solute}} \]
The molecular weight of glucose (C₆H₁₂O₆) can be calculated as follows:
- Carbon (C): 6 atoms × 12 g/mol = 72 g/mol
- Hydrogen (H): 12 atoms × 1 g/mol = 12 g/mol
- Oxygen (O): 6 atoms × 16 g/mol = 96 g/mol

Adding these together gives:
\[ \text{Molecular weight of glucose} = 72 + 12 + 96 = 180 \, \text{g/mol} \]
Now, substituting the values:
\[ \text{Number of moles of glucose} = \frac{10 \, \text{g}}{180 \, \text{g/mol}} = \frac{1}{18} \, \text{mol} \]

### Step 4: Convert the weight of the solvent to kilograms
Since molality is defined as the number of moles of solute per kilogram of solvent, we need to convert the weight of the solvent from grams to kilograms:
\[ \text{Weight of solvent in kg} = \frac{90 \, \text{g}}{1000} = 0.090 \, \text{kg} \]

### Step 5: Calculate the molality
Now, we can calculate the molality (m) using the formula:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}} = \frac{\frac{1}{18} \, \text{mol}}{0.090 \, \text{kg}} \]
Calculating this gives:
\[ \text{Molality} = \frac{1}{18} \times \frac{1000}{90} = \frac{1000}{1620} \approx 0.617 \, \text{mol/kg} \]

### Final Answer
The molality of the glucose solution is approximately **0.617 mol/kg**.

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