* N o 27 m mgs04 boilin owL Question × VG Home | Chegg.com se/ gi?ID 441 568 Sec

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* N o 27 m mgs04 boilin owL Question × VG Home | Chegg.com se/ gi?ID 441 568 SecureID=1381 76673080 engage.com/owl-c/quiz.engine/Question.cgiID-441568SecurelD 13817667308 I /quiz engine/Question engage.com/ OWL Question Status: You must answer 2 of 2 questions correctly in the SAME attempt at this Unit to receive credit for it After ans Chemical Formulas Scientific Notation Periodic Table Tables Match the following aqueous solutions with the appropriate letter from the column on the right [ -A Highest boiling point 1.0.19 mCuBr2 2.0.17 m Mnl2 3.0.27 m Mgso 4 0.48 m Sucrose (nonelectrolyte)D.Lowest boiling point B. Second highest boiling point -C. Third highest boiling point CHECK Learning Resources for Question: 12.5a vis Eflect of lonic Compounds on Freezing Point (Assignment

Explanation / Answer

Q.1: Delta(Tb) = i x Kb(H2O) x m

Kb(H2O) is same for all the solution.

Hence higher is the product of molality and Vanthoff factor( i x m) higher is the elevation in boiling point. Hence higher is the boiling point

For 0.19 m CuBr2 solution: i = 3 ( 1 Cu2+ and 2 Br-)

Hence i x m = 3 x 0.19 = 0.57

For 0.17 m MnI2 solution: i = 3 (1 Mn2+ and 2 I-)

Hence i x m = 3 x 0.17 = 0.51

For 0.27 m MgSO4 solution: i = 2 (1 Mg2+ and 1 SO42-)

Hence i x m = 2 x 0.27 = 0.54

For 0.48 m sucrose(C12H22O11) solution: i = 1 (nonelectrolite)

Hence  i x m = 1 x 0.48 = 0.48

0.19 m CuBr2 ---- highest boiling point(0.57)

0.17 m MnI2 ---- 3rd highest boiling point(0.51)

0.27 m MgSO4 --- second highest boiling point(0.54)

0.48 m sucrose --- lowest boiling point(0.48)

Q.2: Delta(Tb) = i x Kb(H2O) x m

Kb(H2O) is same for all the solution.

Hence higher is the product of molality and Vanthoff factor( i x m) higher is the elevation in boiling point. Hence higher is the boiling point

For 0.20 m BaS solution: i = 2 ( 1 Ba2+ and 1S2-)

Hence i x m = 2 x 0.20 = 0.40

For 0.22 m ZnSO4 solution: i = 2 (1 Zn2+ and 1 SO4^2-))

Hence i x m = 2 x 0.22 = 0.44

For 0.11 m MgBr2 solution: i = 3 (1 Mg2+ and 2 Br-)

Hence i x m = 3 x 0.11= 0.33

For 0.37 m sucrose(C12H22O11) solution: i = 1 (nonelectrolite)

Hence  i x m = 1 x 0.37 = 0.37

0.20 m BaS ---- second boiling point(0.40)

0.22 m ZnSO4 ---- highest boiling point(0.44)

0.11 m MgBr2 --- lowest boiling point(0.33)

0.37 m sucrose --- 3rd highest boiling point(0.37)

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