Question

1 5. What is the vapor pressure of a 3.0 M solution of C.H1206 (aq) (MW = 180.16 g/mol, solution density = 1.12 g/ml) at 100°C? Assume that C6H12O6 is non-volatile . Wil je horen esame here fortion of liter 0. (en) (

Answer 1

5.Ans :-

From Raoult's law fro non-volatile solute :

P_s = P⁰_A.X_A ......................(1)

Where,

P_s = Vapor pressure of solution of glucose

P⁰_A = Vapor pressure of pure solvent i.e. water at 100 ⁰C is = 658 torr

X_A = Mole fraction of solvent = No. of moles of solvent (n_A) / Total no. of moles of solution (n_A+n_B)

Given,

Molarity of C₆H₁₂O₆ = 3.0 M, which means 3.0 moles of C₆H₁₂O₆ (n_B) present in 1 L or 1000 mL of the solution.

So, Volume of the solution = 1000 mL

Given Density of the solution = 1.12 g/mL

Therefore,

Mass of the solution = Volume of the solution x Given Density of the solution = 1000 mL x 1.12 g/mL = 1120 g

Also, Mass of C₆H₁₂O₆ = Moles of C₆H₁₂O₆ x Gram molar mass of C₆H₁₂O₆ = 3.0 mol x 180.16 g/mol = 540.48 g

So, Mass of solvent i.e. water = Mass of solution - Mass of solute

= 1120 g - 540.48 g

= 579.52 g

Therefore, No. of moles of water (n_A) = Mass of water / Gram molar mass of water

= 579.52 g / 18 g/mol

= 32.1956 mol

So, X_A = n_A/(n_A+n_B)

X_A = 32.1956 mol / (32.1956 mol + 3.0 mol) =  32.1956 mol / (35.1956 mol) = 0.915

Put this value in equation (1) :

P_s = 658 torr .(0.915)

P_s = 602.07 torr

Therefore, Vapor pressure of solution of glucose = Ps = 602.07 torr