Question

Calculate the osmotic pressure of solution of 3% solution of urea (m=60) and of 1.5% solution of cane sugar (m=342) at 293K

Answer 1

Osmotic presuure=i*C*R*T,
where, C=concentration of solute(in terms of Molarity)
R= Gas constant=0.082 L(atm)(mol)-1K-1
T=temperature (in Kelvin)
i=Van’t-Hoff factor(=1 for non-electrolyte)
5% urea solution means 5g urea is present in 100ml of solution.

mole of urea=weight given/Molecular weight of urea
= 5g/60gmol-1=1/12
Hence Concentration(C) of urea (in terms of Molarity)
= (mole of urea(n)/Volume of solution)*1000
={(1/12)/100}*1000
=10/12

Hence Osmotic pressure=1*(10/12)*0.082*272 atm
=18.59 atm

Answer 2

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Calculate the osmotic pressure of solution of 3% solution of urea (m=60) and of 1.5% solution of cane sugar (m=342) at 293K

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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