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A solution containing 0.4 g of a non-volatile solute in 0.1 dm3 of the solution exerts an osmotic pressure of 66.66 kPa at 300 K. Thus, molar mass of the solute is
  • a)
     150 g mol-1
  • b)
    300  g mol-1
  • c)
    175  g mol-1
  • d)
    168  g mol-1
Correct answer is option 'A'. Can you explain this answer?
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A solution containing 0.4 g of a non-volatile solute in 0.1 dm3 of the...
Given information:
- Mass of non-volatile solute = 0.4 g
- Volume of solution = 0.1 dm3
- Osmotic pressure = 66.66 kPa
- Temperature = 300 K

To find: Molar mass of the solute

Formula: π = MRTi

Where,
π = Osmotic pressure
M = Molarity of the solution
R = Gas constant = 8.314 J mol-1 K-1
T = Temperature
i = Van't Hoff factor (for non-electrolytes, i = 1)

Calculation:
Molarity of the solution = (0.4 g / Molar mass) / 0.1 dm3
Molarity of the solution = 4 / (Molar mass x 1000)

Putting the values in the formula:

66.66 = (4 / (Molar mass x 1000)) x 8.314 x 300 x 1
Molar mass = 150 g mol-1

Therefore, the molar mass of the solute is 150 g mol-1.

Answer: (a) 150 g mol-1
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A solution containing 0.4 g of a non-volatile solute in 0.1 dm3 of the solution exerts an osmotic pressure of 66.66 kPa at 300 K. Thus, molar mass of the solute isa)150 g mol-1b)300 g mol-1c)175 g mol-1d)168 g mol-1Correct answer is option 'A'. Can you explain this answer?
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