Question

Calculate the osmosis pressure of 5% urea at 270K

Answer 1

We know ,

• Gram Atomic Mass of Urea = 60 g

$\displaystyle\sf{Amount\:of\:Urea=\frac{5}{100}\times 60}$

$\implies{\displaystyle\sf{Amount\:of\:Urea=0.05\times 60}}$

$\implies\displaystyle\sf{Amount\:of\:Urea= 3 g}$

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$\boxed{\displaystyle\sf{Number\:of\:Moles=\frac{Given\:Mass}{GMM}}}$

$\implies\displaystyle\sf{n=\frac{3}{60}}$

$\implies\displaystyle\sf{n=0.05}$

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$\boxed{\displaystyle\sf{Concentration \:of \:Solute=\frac{Moles\:of\:Urea}{Volume \:of \:Solution}\times 1000}}$

Here,

• Moles of Urea = 0.05

• Volume of Solution = 100 ml

Applying the values to the equation ,

$\implies\displaystyle\sf{C=\frac{0.05}{100}\times 1000}$

$\implies\displaystyle\sf{C=0.05\times 10}$

$\implies\displaystyle\sf{C=0.5\:M}$

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The osmotic pressure can be determined from the equation ,

$\boxed{\displaystyle\sf{\pi=i\times C\times R\times T}}$

Where,

• ${\displaystyle\sf{\pi=Osmotic\:Pressure}}$

• $\displaystyle\sf{i=Van't-Hoff\:Factor}$

• $\displaystyle\sf{C=Concentration \:of\:solute\:(in\:terms\:of\:Molarity)}$

• ${\displaystyle\sf{R=Universal\:Gas\:constant}}$

• ${\displaystyle\sf{T=Temperature\:(in\:Kelvin)}}$

Here,

• i = 1

• C = 0.5 M

• R = 0.082 L atm mol⁻¹ K⁻¹

• T = 270 K

Applying the values to the equation ,

$\implies\displaystyle\sf{\pi=1\times 0.5\times 0.082\times 270}$

$\implies\displaystyle\sf{\pi=11.07\:atm}$

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Therefore ,

$\boxed{\bold{\displaystyle\sf{Osmotic\:Pressure=11.07\:atm}}}$