Please show your works and give me finalanswers. In an experiment, 200.0g of a pure solvent is found to have afreezing point of 16.66°C. Its freezing point depressionconstant is known to be 3.90°C/m. Into this solvent 16.92g ofan unknown is dissolved. (Assume that the compound isnot ionic). The new freezing point of thesolution is measured as 14.21°C 1.The molality represents the number of moles of solute (theunknown compound) per 1000g of solvent. What is the number of moles for the 200.0g of solventwhich is present? 2. What is the molecular weight of the unknown compound(grams/mole)? In an experiment, 200.0g of a pure solvent is found to have afreezing point of 16.66°C. Its freezing point depressionconstant is known to be 3.90°C/m. Into this solvent 16.92g ofan unknown is dissolved. (Assume that the compound isnot ionic). The new freezing point of thesolution is measured as 14.21°C 1.The molality represents the number of moles of solute (theunknown compound) per 1000g of solvent. What is the number of moles for the 200.0g of solventwhich is present? 2. What is the molecular weight of the unknown compound(grams/mole)?
Explanation / Answer
mass of solvent m' = 200 g = 0.2Kg Depression in freezing point Tf = freezing point of puresolvent - freezing point of solution = 16.66 oC - 14.21 oC = 2.45 oC freezing point depression constant Kf =3.90°C/m We know that Tf = Kf * m Where m = molality of the solution = No . of moles/ Mass of the solvent in Kg = n / m' So Tf = Kf *(n/m') n = m' *Tf / Kf = 0.2 Kg * 2.45 oC / 3.90 oC / m = 0.1256 mole So the no . of moles for the 200.0g of solventwhich is present is = 0.1256 mol But No . of moles , n = mass / Molar mass 0.1256 = 16.92 g / Molar mass So, molar mass of the unknown M = 16.92 / 0.1256 = 134.7133 g /mol