Question

At 300K 36g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar.If of osmotic pressure of this soluion is 1.52bar at the same temperature what will be its concentration?

Answer 1

pi1=c1RT
pi2=C2RT
Pi1/pi2=C1/C2
Let us calculate the concentration of the first solution with osmotic pressure of 4.98 bar:
Mass of glucose = 36 g
Molar mass of glucose = 180 g/mol
Number of moles of glucose = 36/180 = 0.2 moles Volume of the solution = 1 L Molarity = Number of moles of glucose / Volume of the solution
Molarity = 0.2 moles / L π1= 4.98 bar π1= 1.52 bar C2=1.52*.2/4.98

Answer 2

$\huge\bold{\green{Answer :-}}$

Concentration of the second solution = $\bold{0.061\:M}$

$\fbox{\texttt{\pink{Case\:1\::}}}$

Osmotic pressure = $\bold{4.98\:bar}$

Weight (W) = $\bold{36\:g}$

Volume (V) = $\bold{1\:litre}$

$\fbox{\texttt{\orange{Case\:2\::}}}$

Osmotic pressure = $\bold{1.52\:bar}$

Weight (W) = $\bold{36\:g}$

Volume (V) = $\bold{1\:litre}$

$\fbox{\texttt{\red{For\:case\:1\::}}}$

$\implies\bold{\pi\:V=\frac{W_B}{M_B}\times{RT}}$

$\implies\bold{4.98\times1=\frac{36}{180}\times{R}\times{T}}$

$\implies\bold{4.98=0.2\:RT}\longrightarrow{(1)}$

$\fbox{\texttt{\blue{For\:case\:2\::}}}$

$\implies\bold{1.52={C}\times{R}\times{T}}$

$\implies\bold{1.52=CRT}\longrightarrow{(2)}$

$\bold{\green{Comparing\:equations\:(1)and\:(2),we\:have}}$

$\implies \bold{\frac{4.98}{1.52}=\frac{0.2\:RT}{CRT}}$

$\implies\bold{C=\frac{0.2\times{1.52}}{4.98}=0.061\:M}$