Answer:
Osmotic pressure of the solution will be 818.203 Pa.
Explanation:
Osmotic pressure of a solution is defined by the formula,
π = MRT
where π = Osmotic pressure
M = Molarity of the solution
R = Ideal gas constant
T = Temperature in °K
40 grams of glucose was dissolved in water so the Molarity of the solution will be,
M = [tex]\frac{\text{Number of moles of solute}}{\text{Volume of the solution in liters}}[/tex]
Moles of solute = [tex]\frac{\text{weight of the solute taken}}{\text{volume of the solution}}[/tex]
= [tex]\frac{40}{180}[/tex]
= [tex]\frac{2}{9}[/tex] moles
Morarity = [tex]\frac{\frac{2}{9}}{0.7}[/tex]
= 0.3175 mole per liter
Value of ideal gas constant 'R' = 8.314 [tex]JK^{-1}\text{Mol}^{-1}[/tex]
T = 37°C = (273 + 37)° K
= 310°K
Now by substituting these values in the formula,
π = [tex]0.3175\times 8.314\times 310[/tex]
= 818.203 Pa
Therefore, osmotic pressure of the solution will be 818.203 Pa.
