What is the osmotic pressure of a 5% sucrose (C12H22O11) solution at 10 degrees

Question

What is the osmotic pressure of a 5% sucrose (C12H22O11) solution at 10 degrees Celsius? The density of the solution is 1.035g/mL. To what height could water be pushed up by this pressure? (The density of Hg is 13.6 g/mL, assume the density of water to be 1 g/mL.) 3. What is the osmotic pressure of a 5% sucrose (C12H2O11) solution at 10 degrees Celsius? The density of the solution is 1.035g/mL. To what height could water be pushed up by this pressure? (the density of Hg is 13.6g/mL, assume the density of water to be 1glmL).

Explanation / Answer

Step-1 calculate the molarity of sucrose

Given-

5%solution of sucrose

The density of the solution is 1.035 g/mL

MW of C12H22O11 = 342.2948 g/mol

Solution has a sucrose concentration of 5 %, which means there are 5.0 grams sucrose/100.0 grams of solution.

To get the molarity, convert the above concentration to moles sucrose/Liter solution

5.0 g sucrose/100 x 1.035 g solution/1ml solution x 1000 mL/1L x 1 mol sucrose /342.2948 g sucrose

= 0.151185 mol/L

Step-2

Calculate the osmotic pressure

T= 10 degrees Celsius =283K

R = 0.08206 L-atm/mol-K

O.P. = MRT

Substituting above value in equation-

O.P. = (0.15118 mol/L)(0.08206 L-atm/mol-K)(283 K)

= 3.51085atm

= 3.51atm

Step-3

The height could water be pushed up by this pressure

The density of Hg is 13.6 g/mL

The density of water to be 1 g/mL

O.P. = 3.51 atm x 760 mmHg/1 atm = 2667.6mmHg

O.P. = 2667.6mmHg x 13.6 =36279.36 mmWater

36.28 meters water be pushed up by this pressure

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