If 2 moles of sugar are dissolved in 1000g of water, the elevation in boiling point of the resulting solution is: (Kb = 0.52 K kg mol-1):
0.52° C
100 °C
1.04°C
101.04K
Tb = Kb x m = 0.52 x 2 = 1.04 K or 1.040C
At 100∘C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb=0.52, the boiling point of this solution will be
At 100∘ C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be (Mark nearest one)
The molal boiling point constant for water is 0.513°C kgmol-1. When 0.1 mole of sugar is dissolved in 200g of water, the solution boils under a pressure of 1 atm at: