Question

5 g of non-volatile water soluble compound $X$ is dissolved in 100 g of water. The elevation in boiling point is found to be 0.25. The molecular mass of compound $X$ is

Answer 1

102.4 g mol$^{-1}$

Answer 2

Solution Outline

1. Boiling Point Elevation Relation

   $$\Delta T_b = K_b \times m$$

   where
   $\Delta T_b = 0.25\,^\circ\mathrm{C}$ (observed)
   $K_b = 0.512\,\mathrm{K\cdot kg\,mol^{-1}}$ (for water)

2. Calculate Molality ($m$)

   $$m = \frac{\Delta T_b}{K_b} = \frac{0.25}{0.512} \approx 0.4883\;\mathrm{mol\cdot kg^{-1}}$$

3. Determine Moles of Solute
   Molality is moles of solute per kilogram of solvent.

   $$\text{Mass of water} = 100\;\mathrm{g} = 0.100\;\mathrm{kg}$$ $$\text{Moles of solute} = m \times (\text{kg of solvent}) = 0.4883 \times 0.100 = 0.04883\;\mathrm{mol}$$

4. Find Molar Mass of $X$

   $$M = \frac{\text{Mass of solute}}{\text{Moles of solute}} = \frac{5.000\;\mathrm{g}}{0.04883\;\mathrm{mol}} \approx 102.4\;\mathrm{g\cdot mol^{-1}}$$