Question

Calculate the osmotic pressure of 5% solution of sugarcane ( sucrose) at 15 degree C.

Answer 1

p = {5/342 x 0.1L} x 288 K x 0.0821p = 3.457atm.

Answer 2

The osmotic pressure of 5% solution of sugarcane ( sucrose) at 15 degree C is 3.5 atm

Explanation:

$\pi$=iCRT

$\pi$ = osmotic pressure  = ?

i= vantt hoff factor (1 for non electrolyte)

C= concentration in Molarity

R= solution constant  = 0.0821 Latm/Kmol

T= temperature = $15^0C=(15+273)K=288K$

For the given solution: 5% w/v i.e.  grams of sucrose is dissolved in 100 ml of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

Putting in the values we get:

$C_{sucrose}=\frac{5\times 1000}{342\times 100}=0.15M$

$\pi=0.15mol/L\times 0.0821Latm/Kmol\times 288K$

$\pi=3.5atm$

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