Question

a sample of urine freezes at -0.69C. if you assume that iurine =1 , what is it's osmotic pressure at 25C. with what molar concentration of sucrose would this be isotonic? if you assume iNaCl=1.82, with what molar concentration of NaCl would this be isotonic?

Answer 1

molar depression freezing constant of water(K_f) = 1.86 °C/m

∆T_f = iK_fm

∆T_f = change in freezing point

m = molality

given i = 1

0.69 ^oC = 1 x 1.86 ^oC/m x m

m = 0.371 m m stands for molality

assume density of water = density of urine and mass of solute ~ 0

molality ~ molarity

T = temperature = 25 + 273 = 298 K

Osmotic pressure = pi x i x C x R x T

= 3.14 x 0.371 M x 0.08206 atm.L/mol.K x 298 K

= 28.49 atm

for sucrose i = 1 and hence the isotonic(same osmotic pressure) solution should have same concentration

= 0.371 M

for NaCl

3.14 x 0.371 M x 0.08206 atm.L/mol.K x 298 K = 3.14 x 1.82 x C x 0.08206 atm.L/mol.K x 298 K

C = 0.204 M