if (pie)1 is the osmotic pressure of a solution containing 6g of acetic acid per dm^3 and (pie)2 is that of a solution containing 5.85g of nacl per dm^3 at the same temp then
Answers
The osmotic pressure of the given solution = 10.8 atm.
Explanation:
Osmotic pressure depends on the amount of solute added.
π = i CRT
where π is the osmotic pressure, i is the vanthoff factor 2 for NaCl and C is the concentration in molarity, R is the solution constant 0.0821 Latm/Kmol and T is temperature in Kelvin scale.
For 3.42g sugar in 500 ml
Moles of solute = given mass / molar mass = 3.42 / 342 = 0.01 moles
For 5.85g NaCl in 500 ml
Moles of solute = given mass / molar mass = 5.85 / 58.5 = 0.1 moles
So total moles in the solution = 0.01 + 0.1 = 0.11 moles
Molarity of the solution = 0.11 moles * 1000 / 500 = 0.22 mole/L
π = i CRT
= 2 * 0.22 * 0.0821 * 300
= 10.8 atm.
The osmotic pressure of the solution containing 5.85 g NaCl and 3.42g sugar in 500 ml at 27°C is 10.8 atm.
Given:
Mass of acetic acid, w1 = 6 gm
Mass of NaCl, w2 = 5.85 gm
Volume of both solutions = 1 dm³ = 1000 ml
To Find:
The relation between the osmotic pressure of both the solutions.
Calculation:
- Molarity of acetic acid = (w1 × 1000)/(M.wt1 × V)
⇒ M1 = (6 × 1000)/(60 × 1000)
⇒ M1 = 0.1 M
- Molarity of NaCl = (w2 × 1000)/(M.wt2 × V)
⇒ M1 = (5.85 × 1000)/(58.5 × 1000)
⇒ M2 = 0.1 M
- The osmotic pressure of acetic acid solution:
Π1 = M1RT ...(1)
- The osmotic pressure of NaCl solution:
Π2 = M2RT ...(2)
- Divide eqn (1) by eqn (2):
Π1/Π2 = M1RT/M2RT
⇒ Π1/Π2 = M1/M2
⇒ Π1/Π2 = 0.1/0.1