An analytical laboratory prepares a aqueous solution of Sodium thiosulphate (). The density of this — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
An analytical laboratory prepares a $3.0\text{ M}$ aqueous solution of Sodium thiosulphate ($Na_2S_2O_3$). The density of this solution is carefully measured to be $1.25\text{ g/mL}$. Which of the following analytical statements regarding this specific solution are mathematically correct? (Atomic weights: Na=23, S=32, O=16)
💡 Solution & Explanation
Step 1: Base values for 1 Litre ($1000\text{ mL}$) of solution. Molar mass of $Na_2S_2O_3 = (23\times 2) + (32\times 2) + (16\times 3) = 158\text{ g/mol}$. Mass of 1 L solution = $1000\text{ mL} \times 1.25\text{ g/mL} = 1250\text{ g}$. Mass of solute in 1 L = $3.0\text{ moles} \times 158\text{ g/mol} = 474\text{ g}$. Mass of solvent in 1 L = $1250 - 474 = 776\text{ g}$ (Statement D is true). Step 2: Mass % = $(474 / 1250) \times 100 = 37.92\%$ (Statement A is true). Step 3: Molality ($m$) = Moles solute / Mass solvent in kg = $3.0 / 0.776 = 3.865\text{ m}$ (Statement C is true). Step 4: Mole fraction = Moles solute / Total moles. Moles of water = $776 / 18 = 43.11\text{ moles}$. Mole fraction = $3.0 / (3.0 + 43.11) = 3.0 / 46.11 = 0.065$ (Statement B is true).