Molality of 10% (w/w) aq. glucose solution is

Options

A. `(25)/(81)m`
B. `(10)/(9)m`
C. `(25)/(9)m`
D. `(50)/(81)m`

Answer

To find the molality of a 10% (w/w) aqueous glucose solution, we can follow these steps:

### Step 1: Understand the Given Information
We have a 10% (w/w) glucose solution, which means that in 100 grams of the solution, there are 10 grams of glucose (solute) and 90 grams of water (solvent).

### Step 2: Identify the Mass of Solute and Solvent
- Mass of solute (glucose) = 10 grams
- Mass of solution = 100 grams
- Mass of solvent (water) = Mass of solution - Mass of solute = 100 g - 10 g = 90 grams

### Step 3: Convert the Mass of Solvent to Kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of the solvent from grams to kilograms:
- Mass of solvent in kilograms = 90 grams × (1 kg / 1000 g) = 0.090 kg

### Step 4: Calculate the Moles of Solute
To find the moles of glucose (C6H12O6), we use its molar mass:
- Molar mass of glucose = (12 × 6) + (1 × 12) + (16 × 6) = 72 + 12 + 96 = 180 g/mol

Now, we can calculate the moles of glucose:
- Moles of glucose = Mass of solute / Molar mass = 10 g / 180 g/mol = 1/18 mol

### Step 5: Calculate the Molality
Now we can calculate the molality (m):
- Molality (m) = Moles of solute / Mass of solvent (kg)
- Molality (m) = (1/18 mol) / (0.090 kg)

To simplify this:
- Molality (m) = (1/18) / (0.090) = (1/18) × (1000/90) = (1000 / 1620) = 50 / 81 molal

### Final Answer
The molality of the 10% (w/w) aqueous glucose solution is **50/81 molal**.

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