If 18 g of glucose is present in 1000 g of solvent, the solution is said to be

Options

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

Answer

To determine the concentration of the solution when 18 g of glucose is dissolved in 1000 g of solvent, we will calculate the molality of the solution step by step.

### Step-by-Step Solution:

1. **Identify the mass of the solute (glucose)**:
- Given: Mass of glucose (solute) = 18 g

2. **Determine the molar mass of glucose**:
- The molar mass of glucose (C6H12O6) is approximately 180 g/mol.

3. **Calculate the number of moles of glucose**:
- Use the formula:
\[ \text{Number of moles} = \frac{\text{Mass of solute}}{\text{Molar mass of solute}} \]
- Substituting the values:
\[ \text{Number of moles} = \frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ moles} \]

4. **Convert the mass of the solvent from grams to kilograms**:
- Given: Mass of solvent = 1000 g
- Convert to kg:
\[ \text{Mass of solvent in kg} = \frac{1000 \text{ g}}{1000} = 1 \text{ kg} \]

5. **Calculate the molality of the solution**:
- Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
- Substituting the values:
\[ \text{Molality} = \frac{0.1 \text{ moles}}{1 \text{ kg}} = 0.1 \text{ molal} \]

6. **Conclusion**:
- The solution is said to be **0.1 molal**.