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Calculate the osmotic pressure of 5% solution of urea at 272K

Calculate the osmotic pressure of 5% solution of urea at 272K

Grade:Upto college level

3 Answers

Naveen Kumar
askIITians Faculty 60 Points
9 years ago
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

ZEESHAN
21 Points
6 years ago
If 3% urea solution is mixed with 1.5% sucrose solution at 278 K what could be the osmotic pressure urea =60 and sucrouce =342
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago

Dear student,
Please find the attached explanation to your question.

Osmotic pressure=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.
Hence assuming 100ml of sample we will have 5g urea
Hence, moles 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

Hope it helps.
Thanks and regards,
Kushagra

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