Please show your works and give me final answers. In an experiment, 200.0g of a
ID: 691369 • Letter: P
Question
Please show your works and give me final answers. In an experiment, 200.0g of a pure solvent is found to have afreezing point of 16.66?. Its freezing point depressionconstant is known to be 3.90?/m. Into this solvent 16.92gof an unknown is dissolved. (Assume that the compound isnot ionic). The new freezing point of thesolution is measured as 14.21? a) What is the freezing point depressionTf and molality(m)of the solution (show your work)? b) The molality represents the number of moles of solute (theunknown compound) per 1000g of solvent. What is the number of molesfor the 200.0g of solvent which is present?( show your work) c) What is the molecular weight of the unknown compound(crams/mole)?(show your work) Please show your works and give me final answers. In an experiment, 200.0g of a pure solvent is found to have afreezing point of 16.66?. Its freezing point depressionconstant is known to be 3.90?/m. Into this solvent 16.92gof an unknown is dissolved. (Assume that the compound isnot ionic). The new freezing point of thesolution is measured as 14.21? a) What is the freezing point depressionTf and molality(m)of the solution (show your work)? b) The molality represents the number of moles of solute (theunknown compound) per 1000g of solvent. What is the number of molesfor the 200.0g of solvent which is present?( show your work) c) What is the molecular weight of the unknown compound(crams/mole)?(show your work)Explanation / Answer
When dealing with freezing point depression the formula you use themost often is this one: Tf = Kf *cm Where Tf is the freezing point depression (i.e.the difference in freezing temperature between the pure substanceand the solution); Kf is the freezing point constant (inoC/m, in this instance); and cm is the molalconcentration (i.e. molality = moles of solute / kg of solvent) ofthe solution. 1. Freezing point depression (Tf) = Tf,pure solvent - Tf, solution = 16.66 - 14.21= 2.45 oC. Tf =Kf * cm cm = Tf /Kf = 2.45oC / 3.90oCm-1 = 0.628 m 2. According to the definition of molality given in the question: cm = moles solute / 1 kg solvent = 0.628 m = 0.628mol / kg solvent moles solute (per kg solvent) = 0.628 mol/kg solvent * 1 kg solvent = 0.628 mol moles solute (per 0.2 kg solvent) = 0.628 mol / 5 = 0.126mol (I think that's right...) 3. Molar mass = mass / moles = 16.92 g / 0.126 mol = 134.7 g/mol