Answer:
The osmotic pressure is 0.4808 atm
Explanation:
Given:
Temperature = [tex]20^{\circ} Celsius[/tex]
Volume of sucrose = 0.69 grams
volume of water = 100 mL
To Find:
The osmotic pressure =?
Solution:
Step 1: lets find the number of moles in 0.69 grams of sucrose([tex]C_{12}H_{22}O_{11})[/tex]
number of moles in 0.69 grams of sucrose = [tex]\frac{1 mol}{342 grams} \times 0.69 g[/tex]
=>[tex]\frac{1 mol}{342 grams} \times 0.69 g[/tex]
=>[tex] 0.0029 \times 0.69 g[/tex]
=>0.002 moles
Step2 :Dividing number of moles by the amount of litres in the solution
=>[tex]\frac{0.002}{0.100}[/tex] [ 100 mL = 0.100L]
=> 0.02 M
Step 3: Converting the temperature to kelvin
=>20 +273
=>293 K
Step 4: Finding the value of osmotic pressure
[tex]\text{Osmotic pressure} \pi = MRT[/tex]
where
M is the Molarity
R is gas constant
T is the temperature
Now substituting the values,
[tex]\text{Osmotic pressure} \pi = 0.02 \times 0.08206 \ times 293[/tex]
[tex]\text{Osmotic pressure} \pi = 0.4808 atm[/tex]