The number of moles of water in 100 g =10018=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
P0−PP0=X
760−732760=nn+5.555
0.0368=nn+5.555
27.144=n+5.555
n=5.55526.14=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=0.2120.1=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
Hence, the correct option is A