Question

The osmotic pressure of a solution is calculated using the formula Π=MRT where Π is the osmotic pressure in atm, M is the molarity, R is the ideal gas constant, and T is the kelvin temperature. Part A What is the osmotic pressure of a solution made by dissolving 40.0 g of glucose, C6H12O6, in enough water to form 700.0 mL of solution at 37.0 ∘C ? Express your answer to three significant figures and include the appropriate units. nothing nothing

Answer 1

Osmotic pressure of the solution will be 818.203 Pa.

Step-by-step 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 =
`\frac{\text{Number of moles of solute}}{\text{Volume of the solution in liters}}`

Moles of solute =
`\frac{\text{weight of the solute taken}}{\text{volume of the solution}}`

=
`(40)/(180)`

=
`(2)/(9)` moles

Morarity =
`((2)/(9))/(0.7)`

= 0.3175 mole per liter

Value of ideal gas constant 'R' = 8.314
`JK^(-1)\text{Mol}^(-1)`

T = 37°C = (273 + 37)° K

= 310°K

Now by substituting these values in the formula,

π =
`0.3175* 8.314* 310`

= 818.203 Pa

Therefore, osmotic pressure of the solution will be 818.203 Pa.