A quantity of 6.4 g sulphur dissolved in 200 g of raises the boiling point of the solvent by 0.32°C. — Solutions and Colligative Properties Chemistry Question
Question
A quantity of 6.4 g sulphur dissolved in 200 g of $CS_2$ raises the boiling point of the solvent by 0.32°C. Boiling point of $CS_2 = 47^\circ\text{C}$ and latent heat of vaporization of $CS_2 = 80.0 \text{ cal/g}$. The molecular formula of sulphur in this liquid is
💡 Solution & Explanation
Calculate ebullioscopic constant: $K_b = (R T_b^2) / (1000 L_{vap}) = (2 \times (273+47)^2) / (1000 \times 80) = (2 \times 320^2) / 80000 = 204800 / 80000 = 2.56 \text{ K kg mol}^{-1}$. Using $\Delta T_b = K_b \times m \Rightarrow 0.32 = 2.56 \times m \Rightarrow m = 0.32 / 2.56 = 0.125$ m. Moles $= 0.125 \times 0.200 \text{ kg} = 0.025$ mol. Molar mass $M = 6.4 \text{ g} / 0.025 \text{ mol} = 256$ g/mol. Since atomic mass of S is 32, number of atoms $= 256 / 32 = 8$. Formula is $S_8$. Therefore, correct answer is A.