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Answer 1

Explanation:

The number of moles of water in 100 g  =18100=5.555

Let n be the number of moles of solute.

The relative lowering in the vapour pressure is equal to the mole fraction of solute

 P0P0−P=X

 760760−732=n+5.555n

 0.0368=n+5.555n

 27.144=n+5.555

 n=26.145.555=0.212

The molality of the solution is the number of moles of solute in 1 kg of water

100 g of water corresponds to 0.1 kg 

 m=0.10.212=2.12

The elevation in the boiling point is 

 ΔTb=kbm=0.52×2.12=1.1oC

The boiling point of the solution is 

 Tb=100+1.1=101.1oC≃101oC

Answer 2

Explanation:

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