Concentration Terms

Question

The molality of a solution prepared by dissolving 9.0 g of glucose (C₆H₁₂O₆, molar mass = 180 g mol⁻¹) in 500 g of water is:

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1. **Recall the Definition of Molality**: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$.
2. **Calculate Moles of Glucose**:
   - Moles = $\frac{9.0 	ext{ g}}{180 	ext{ g/mol}} = 0.05$ mol.
3. **Convert Solvent Mass to kg**:
   - Mass of water = $500 	ext{ g} = 0.500 	ext{ kg}$.
4. **Compute Molality**:
   - $m = \frac{0.05 	ext{ mol}}{0.500 	ext{ kg}} = 0.10 	ext{ mol kg}^{-1}$.
5. **Key Point**: Molality depends on the mass of solvent, not the total mass of solution, making it temperature-independent — an important distinction from molarity for NEET.

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