Question

The osmotic pressure of an aqueous solution at 300K is 250.0 kPa. (1)Calculate the molarity of solute (impurity) M (mol/m³)=; (2) Calculate the molality of solute (impurity) b (mol/kg)=; Calculate the freezing point of the solution T(^oC)=. Please pay attention to the sign. The cryoscopicconstant Kf=1.86Kkgmol^-1. Please enter your answers with 2 decimals. for example, 3.02466 is written as 3.02.

Answer 1

osmotic pressure(P) = CRT

P = 250 kpa = 2.467 atm

C = concentration of aqueous solution = ?

R = 0.0821 l.atm.k-1.mol-1.

T = 300k

2.467 = c*0.0821*300

C = 0.10 M

molarity(M) = 0.1 mol/L

1 L = 10^-3 m^3

Molarity(M) = 0.1/10^-3 = 100.00 mol/m3

density of aqueous solution = 1 g/ml

molality(m) = M*1000/((d*1000)-(M*Mwt))

            = 0.1*1000/((1*1000)-(0.1*18))

            = 0.10 mol/kg

DTf = kf*m

(0-x) = 1.86*0.1

x =freezing point of aqueous solution = -0.19 C