SolutionHard
Question
n aqueous solution of NaCl shows the An aqueous solution point of water equal to 0.372 K. The boiling point of BaCl2 solution of same molarity will be
[Kf(H2O) = 1.86K kg mol-1; Kb(H2O) = 0.52K kg mol-1]
[Kf(H2O) = 1.86K kg mol-1; Kb(H2O) = 0.52K kg mol-1]
Options
A.100.52oC
B.100.104oC
C.101.56oC
D.100.156oC
Solution
ᐃTf = 0.372. Hence, ᐃTb = 0.156oC
i.e., Tb = 100.156oC
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