Question

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

Answer 1

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.

Answer 2

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

⇒ Π1 = Π2

- For the given condition, the osmotic pressure of both the solutions is same, i.e., Π1 = Π2.