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(Solved): Calculate the boiling point of a water solution containing 6.08 M Na2SO4, a strong electrolyte. Firs ...
Calculate the boiling point of a water solution containing 6.08 M Na2SO4, a strong electrolyte. First, What is the n value (number of ions produced from Na2SO4)? (1, 2, 3, 4, 5) Second, Atemp = n(K)M, where Kb = 0.52 °C/M = Third, What is the new solution Boiling point = Hint: Boiling point of water is 100 °C. n value = 2 Atemp = 2.86 °C New Boiling point = 2.86 °C n value = 1 Atemp = 1.43 °C New Boiling point = 101.43 °C n value = 3 Atemp = 9.49 °C New Boiling point = 282.54 °C n value = 1 Atempo = 1.43 °C New Boiling point = 274.43 °C n value = 142.04 Atemp = 203.1 °C New Boiling point = 303.1 °C n value = 3 Atemp = 9.49 °C New Boiling point = 109.49 °C °C
ond, Atempe =n(Kb)M, where Kb=0.52∘CrM= rd, What is the new solution Bolling point = Botiling point of water is 100∘C. n value =2Aterap=2,B6∘C New Bowing point =2.86CC n vatue =1Δtemphb=1.43cc New Bollins point =101.49gC n value =3Δtemps=9.49.C New Bolling point =28254cc h value =1Δstrp0=1.439C New Boilinz point =274.43∘C A value -14204 Atemph =2031cc New fhoiling point = 30s:1 1C. n value =3Δtempb+9.499C