SolutionHard
Question
1.0 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of the solution is ( Kb for H2O = 0.52 K kg/mol)
Options
A.375.5 K
B.374.04 K
C.377.12 K
D.373.25 K
Solution
x3y2 ⇋ 3x2+ + 2y3- for complete ionization.
1 - α nα mα
i = 1 + (m + n - 1) α
i = 1 + (2 + 3 - 1) × 0.25 = 1 + 1 = 2
ᐃTb = i × kb × m = 2 × 0.52 × 1 = 1.04
B.P. of solution (Tb) = ᐃTb + Tbo = 1.04 + 373 = 374.04 K Ans.
1 - α nα mα
i = 1 + (m + n - 1) α
i = 1 + (2 + 3 - 1) × 0.25 = 1 + 1 = 2
ᐃTb = i × kb × m = 2 × 0.52 × 1 = 1.04
B.P. of solution (Tb) = ᐃTb + Tbo = 1.04 + 373 = 374.04 K Ans.
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