Answer:
38.35 bar
Explanation:
We are given that
Temperature=T=36.6 degree Celsius=36.6+273=309.6 K
Given mass of glucose=9.18 g
Molar mass of glucose([tex]C_6H_{12}O_6=6(12)+12(1)+6(16)[/tex]=180 g
Mass of c=12 g,mass of hydrogen=1 g, mass of O=16 g
Volume of solution=34.2 mL
Molarity of solution=[tex]\frac{given\;mass}{molar\;mass\times volume}\times 1000[/tex]
Where volume (in mL)
Molarity of solution=[tex]\frac{9.18}{180\times 34.2}\times 1000=1.49 M[/tex]
We know that
Osmotic pressure=[tex]\pi=MRT[/tex]
Where M=Molarity of solution
R=Constant=0.08314 Lbar/mol k
T=Temperature in kelvin
Using the formula
[tex]\pi=1.49\times 0.08314\times 309.6=38.35 bar[/tex]
Hence, the osmotic pressure=38.35 bar
