What is the boiling point of 1 molal aqueous solution of nacl?
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Answer:
delta Tb=Tb-Tb°
=kb×m×i
Tb°=373K (constant)
Tb-373=0.52×1×2
Tb=1.04+373
Tb=374.04
Hence boiling point of 1molal aqueous solution of nacl is 374.04
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