The molality of the solution containing 70g of glucose dissolved in 1kg of water is ?

Answer

To find the molality of the solution containing 70g of glucose dissolved in 1kg of water, we can follow these steps:

### Step 1: Calculate the moles of glucose
To find the moles of glucose, we use the formula:

\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]

Given:
- Mass of glucose = 70 g
- Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol

Now, substituting the values:

\[ \text{Moles of glucose} = \frac{70 \, \text{g}}{180 \, \text{g/mol}} \approx 0.3889 \, \text{moles} \]

### Step 2: Determine the mass of the solvent in kg
The mass of the solvent (water) is given as 1 kg.

### Step 3: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is:

\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]

Substituting the values we found:

\[ \text{Molality} = \frac{0.3889 \, \text{moles}}{1 \, \text{kg}} = 0.3889 \, \text{mol/kg} \]

### Step 4: Round the answer
For practical purposes, we can round the answer to two decimal places:

\[ \text{Molality} \approx 0.39 \, \text{mol/kg} \]

### Final Answer:
The molality of the solution is approximately **0.39 mol/kg**.

---