Chemistry, asked by rohitgill106, 8 months ago

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

Answered by surender100
0

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

Answered by veenavg78
1

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.

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