Molality of 20% (w/w) aq. glucose solution is

Options

A. `(25)/(18)m`
B. `(10)/(9)m`
C. `(25)/(9)m`
D. `(5)/(18)m`

Answer

To calculate the molality of a 20% (w/w) aqueous glucose solution, follow these steps:

### Step 1: Understand the given information
We have a 20% (w/w) glucose solution, which means that in 100 grams of the solution, there are 20 grams of glucose (solute) and 80 grams of water (solvent).

### Step 2: Identify the mass of solute and solvent
- Mass of glucose (solute) = 20 grams
- Mass of solution = 100 grams
- Therefore, mass of solvent (water) = Mass of solution - Mass of solute = 100 g - 20 g = 80 grams

### Step 3: Calculate the number of moles of glucose
To find the number of moles of glucose, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]
The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol. Thus:
\[ \text{Number of moles of glucose} = \frac{20 \text{ g}}{180 \text{ g/mol}} = \frac{1}{9} \text{ moles} \]

### Step 4: Calculate the mass of solvent in kilograms
Since molality (m) is defined as the number of moles of solute per kilogram of solvent, we need to convert the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent} = 80 \text{ g} = 0.080 \text{ kg} \]

### Step 5: Calculate the molality
Now we can calculate the molality using the formula:
\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}} \]
Substituting the values we have:
\[ \text{Molality} = \frac{\frac{1}{9} \text{ moles}}{0.080 \text{ kg}} = \frac{1}{9 \times 0.080} = \frac{1}{0.72} \approx 1.39 \text{ mol/kg} \]

### Final Answer
The molality of the 20% (w/w) aqueous glucose solution is approximately **1.39 mol/kg**.

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