If freezing point depression constant (K fp ) ofwater is -1.86 C/m, at what temp

Question

If freezing point depression constant (Kfp) ofwater is -1.86 C/m, at what temperature will asolution prepared by dissolving 200. g ofNa2SO4 (molar mass 142.05 g/mol) into 1.00 kgof water freeze (assuming no ion pairing in the solution)? The answer is -7.86 degrees C. Please show how it isso! Thank you! If freezing point depression constant (Kfp) ofwater is -1.86 C/m, at what temperature will asolution prepared by dissolving 200. g ofNa2SO4 (molar mass 142.05 g/mol) into 1.00 kgof water freeze (assuming no ion pairing in the solution)? The answer is -7.86 degrees C. Please show how it isso! Thank you!

Explanation / Answer

Formula : t = Kf *molality molality of given solution =(200g/(142.05g/mol))/1.0kg                                        =1.41m If there is no ion pairing, molality of the soluttion is3*1.41= 4.23 m                        t = -1.86 C/m *  4.23 m                                = -7.86780C   .............>I      We know taht, freezingpoint of solvent - freezing point of solution = t=  -7.86780C                                  freezingpoint of solution = 00C + 7.86780C                                                                        =7.86780C      We know taht, freezingpoint of solvent - freezing point of solution = t=  -7.86780C                                  freezingpoint of solution = 00C + 7.86780C                                                                        =7.86780C

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.