the osmotic pressure of solution of non volatile , non -electrolyte solute is 7.6 atm at 27 degree celcius , then boiling point of same solution is (kb and molarity same)

Dear Student,

Please find below the solution of your asked query:

For finding the boiling point of the solution, we will firstly calculate the concentration by using the formula:

π=mRT
m = molality (concentration)
R = 0.0821 L atm mol-1 K-1 (Gas constant)
T = 27o C = 300 K   (Temperature)
π = 7.6 atm  (Osmotic pressure)

7.6=m×0.0821×300m=7.60.0821×300m=7.624.63m=0.31 mol/kg

Now, we will calculate the Elevation in freezing point by using the relation:

Tb=Kbm       (Since it is a non-electrolyte solute)
Kb = 0.512 °C/molal     (Since, no information is given about Kb, so I am assuming that we have water as a solvent & this is the boiling point constant of water)
m = 0.31 mol/kg

So,

Tb=0.512×0.31Tb=0.16 oC

Now, finally the boiling point will be,

Tb=Tb-Tbo0.16=Tb-100Tb=100+0.16Tb=100.16 oC

Hope it Helps!!

Regards

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