Answer:
The vapour pressure of a pure solvent decrease .
when a non-volatile solute is added to the solvent
this is because on adding the solute a fewer number of water molecules are present at the surface which can evaporate as some of the area is occupied by -non- volatile solute molecules thereby decreasing the vapour pressure of the solution of the glucose in wateris lower than pure water.
Answer;
Let the density of solution = d g/cm3
and volume of solution is = 1L = 1000 cm3
mass of the solution = d xV
= (1000d)g
6.90 M solution mean 1L solution contains 6.90 moles of KOH
therefore mass of KOH = 6.90 x 56 = 386.4g
but only 30% of the solutionby mass is KOH
therefore
30/100 x (1000)d =386.4
and density is 1.288g/ cm3
Answer:
The molality of the cane sugar, m= 0.1539m
depression in freezing point = 273.15-271
=2.15K
Since = Kfm
or Kf = /m = 2.15k/0.1539m = 13.97K/m
Now the weight of glucose , W2 = 5g
molecular mass of glucose, M2 =180g/mol
then =Kfm
=
then freezing point of solution = 273.15-3.88
=263.27K
An aqueous solution of 2 percent non-volatile solute exerts a pressure of 0.096 atm at the boiling poine of the solvent. What is the molecular mass of the solute ?