Arrange the following aqueous solutions in order of increasing freezing point: 0

Question

Arrange the following aqueous solutions in order of increasing freezing point: 0.10 m urea, 0.06 m NaCI, 0.05 m Ba(N0_3)_2, and 0.060 m sucrose. Calculate the freezing point of each solution, given that pure water has a normal freezing point of 0.00 degreeC and a molal freezing-point depression constant (K_f) is 1.86 degreeC/m. 6. An isotonic saline solution, which has the same osmotic pressure as blood in the human body, can be prepared by dissolving 1.615 g of NaCl in enough water to produce 175.0 uiL of solution. What is the osmotic pressure, in mm Hg, of this solution at 25.0 degreeC? Use the experimentally determine van't Hoff factor of 1.80 for NaCl at this concentration.

Explanation / Answer

Depression in freezing point= kf*m*i

i= van't Hoff factor

for Urea i=1

depression in freezing point= 1.86*0.10*1=0.186

Freezing point depression = Freezing of pure solvent- freezing point of solution

Freezing point of solution= 0 -0.186= -0.186 deg.c

for NaCl Van;t Hoff factor= 1 (for Na+) +1 (Cl-) =2

for 0.06m NaCl, Freezing point = 0- 1.86*0.06*2 =-0.2232 deg.c

for Ba(NO3)2 i= 3

freezing point = 0- 0.186*0.05*3=-0.0279 deg.c

for sucrose i=1 ( neither associates nor dissociates)

freezinng point= 0- (0.186*0.06)= -0.01116 deg.c

So the order is NaCl, Urea, Ba(NO3)2, Sucrose

6,

Osmotic pressure = CRT

C= concentration of NaCl= moles/L

moles of NaCl= mass/moelcualr weight= 1.615/58.5=0.0276 moles

Concentration = molea/Liter of solution = 0.0276*1000/175=0.1577M

Osmotic pressure =Van't Hoff factor (1.8) * 0.1577*0.08206 L.atm/mole.K *(25+273.15)=6.94 atm

Osmotic pressure in mmHG= 6.94*760=5274.4 mm Hg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.