Question

18g of glucose is dissolved in 1 kg of water in a saucepan. at what temperature. will the water boil at 1.013bar pressure? kb for water is 0.52kkg/mol ​

Answer 1

Wt of glucose=18 g

Mol wt glucose= 180 g

Moles of glucose= 18/180= 0.1 mole

Weight of solvent = 1 kg

∆Tb = Kb × m

m = moles of solute / weight of solvent

m= 0.1/ 1

∆Tb = 0.52×0.1=0.052

As the temperature is (Tb•) 373 K at 1.013 bar

Tb•-Tb = ∆Tb

Tb= 373+ 0.052= 373.052

I hope this helps( ╹▽╹ )