Calculate the boiling and freezing points of 0.221 mol/kg aqueous solutions of NaCl and of CaCl,.К,(Н-0) %3 0.520 °C/mK{(H2O) = 1.86 °C/mТь (Н-0) — 100.00 °CTf (H2O) = 0.00 °CT(NaCl) solution:°CТЬ(СаCl>) solution:°CT{(NaCl) solution:°CT(CaCl2) solution:°CSubmitShow Approach Show Tutor Steps

Given data,molality of NaCl=0.221mol/Kgmolality of CaCl2=0.221mol/KgKb(H2O)=0.520oC/mKf(H2O)=1.86oC/mTb(H2O)=100oCTf(H2O)=0.00oCFor NaCl :- NaCl→Na++Cl-i=2Using,∆Tb=i×Kb×mwhere, i=van't hoff factor,m=molality∆Tb=2×0.520oC/m×0.221mTb(NaCl)-100oC=0.2298oCTb(NaCl)=0.2298oC+100oC=100.2298=100.23oCUsing,0.0oC-Tf(NaCl)=i×Kf×m0.0oC-Tf(NaCl)=2×1.86oC/m×0.221m0.0oC-Tf(NaCl)=0.822oCTf(NaCl)=0.00-0.822=-0.822oCFor CaCl2 :- CaCl2→Ca2++2Cl-i=3Using,∆Tb=i×Kb×mTb(CaCl2)-100oC=3×0.520oC/m×0.221mTb(CaCl2)=0.3447+100=100.3447oCUsing,∆Tf=i×Kf×m0.0oC-Tf(CaCl2)=3×1.86oC/m×0.221m0.0oC-Tf(CaCl2)=1.23oCTf(CaCl2)=-1.23oC

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Calculate the boiling and freezing points of 0.221 mol/kg aqueous solutions of NaCl and of CaCl2. К,Н,0) — 0.520 °C/m K(H2O) = 1.86 °C/m Ть (Н-0) %3D 100.00 °C Tf (H2O) = 0.00 °C T,(NaCl) solution: °C Tp(CaCl,) solution: °C T(NaCl) solution: °C T(CaCl,) solution: °C

Calculate the boiling and freezing points of 0.221 mol/kg aqueous solutions of NaCl and of CaCl2. К,Н,0) — 0.520 °C/m K(H2O) = 1.86 °C/m Ть (Н-0) %3D 100.00 °C Tf (H2O) = 0.00 °C T,(NaCl) solution: °C Tp(CaCl,) solution: °C T(NaCl) solution: °C T(CaCl,) solution: °C

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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