Question

For a 5% solution of urea (Molar mass = 60 g/mol), calculate the osmotic
pressure at 300 K. (R = 0·0821 L atm K-1 mol-1
​

Answer 1

Answer:

Explanation:

Osmotic presuure=i∗C∗R∗T

where, C=concentration of solute(in terms of Molarity)

R= Gas constant =0.082L(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

=5g60gmol−1=112

Hence Concentration of urea

(in terms of Molarity)

= (moles of urea(n) /volume of solution ) ×1000

={(112)/100}∗1000

=1012

Hence Osmotic pressure =1×(10/20)×0.082×300 atm

=20.52atm

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Answer 2

The osmotic is $20.52atm$

Given-$5$% solution of urea

molar mass$=60g/mol$

Find the calculate the osmotic pressure

Solution-In chemistry, the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles. The molar mass is a bulk, not molecular, property of a substance.

$5$% Urea solution means $5g$ Urea is present in $100ml$ of solution. Moles of urea present $=\frac{weight of given}{molecular weight of urea}$

$=59/60g/mol$

$=112$

Here the concentration of urea$=\frac{moles of urea}{volume of solution}\times 1000$

$=\frac{112}{100}\times 1000$

$=1012$

Here osmotic pressure $=1\times \frac{10}{20} \times 0.082\times 300$

$=20.52atm$

Hence, the osmotic pressure is $20.52atm$

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