At 100`C the vapour pressure of a solution of 6.5g of a solute in 100g water is 732mm.if Kb=0.52,the boiling point of this solution will be
Dear Student ,
We know that relative lowering of vapour pressure is given by the formula
P0A of pure water = 760 mm Hg
PA of the solution = 732 mm Hg
This implies = 0.037
Now as we have got the mole fraction of the solute , we can find the no. of moles of solute present
0.037 =
This gives , nsolute = 0.213
Hence we have 0.213 moles of the solute present . Now we know the formula for elevation in boiling point.
ΔTb = kb * m =
Hence ΔTb = 1.1076
Therefore the boiling point will be = (100 + 1.1076) = 101.1076o C
Hence the boiling point of the solution will be 101.1076o C.
I hope you understood . Keep posting.
We know that relative lowering of vapour pressure is given by the formula
P0A of pure water = 760 mm Hg
PA of the solution = 732 mm Hg
This implies = 0.037
Now as we have got the mole fraction of the solute , we can find the no. of moles of solute present
0.037 =
This gives , nsolute = 0.213
Hence we have 0.213 moles of the solute present . Now we know the formula for elevation in boiling point.
ΔTb = kb * m =
Hence ΔTb = 1.1076
Therefore the boiling point will be = (100 + 1.1076) = 101.1076o C
Hence the boiling point of the solution will be 101.1076o C.
I hope you understood . Keep posting.