calculate the osmotic pressure of solution obtained by mixing 100 ml of 3.4 % solution of urea and 100 ml of 1.6 % solution of cane sugar at 293 Kelvin
Answers
Answer:
Explanation:
Urea and cane sugar are very dissolvable in water. But we assume these molecules cannot pass through a semipermeable membrane. Only water can pass through.
Π = Osmotic pressure = C_solute * R T
R = 8.314 J/°K
C_solutes = Molar concentration of solutes
T = absolute temperature = 293 °K given
We assume that the volume of Urea and Cane sugar in the solutions are negligible as the concentrations are low.
Molar concentration: in 200 ml we have
Solutes = 3.4 / 60 moles of Urea + 1.6 /342 moles of cane sugar
= 0.061 moles
= 0.061 * 1000ml /200 ml = 0.306 M
Since Cane sugar and Urea are not Ionic, we do not multiply the concentration by any factor.
Osmotic pressure = Π = 0.306 * 8.314 * 293
= 745 Pa
Answer:
The number of moles of urea present =100mL×
100mL
3.4g
×
60g
1mol
=0.0566 mol.
The number of moles of sucrose present =50mL×
100mL
1.6g
×
342g
1mol
=0.002339 mol.
Total number of moles n=0.0566mol+0.002339mol=0.0589mol
Total molar concentration is
100mL+50mL
0.0589mol
×
1L
1000mL
=0.3929 M
The osmotic pressure is Π=CRT=0.3929 M×0.08206 Latm−mol/K×300.15 K=9.704 atm.