Osmotic pressure

Chemistry Level 2

A chemist dissolves 2.00 grams of protein in 0.100 liters of water. The osmotic pressure is 0.042 atm at 298 K. What is the approximate molar mass of the protein?

1 , 150 g mol − 1 1,150\text{ g mol}^{-1} 4 , 600 g mol − 1 4,600\text{ g mol}^{-1} 11 , 500 g mol − 1 11,500\text{ g mol}^{-1} 23 , 000 g mol − 1 23,000\text{ g mol}^{-1}

Krishna Singh by Krishna Singh , 22, India

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3 solutions

Pranjal Jain
Oct 31, 2014

π = C R T ⇒ π = n V R T ⇒ π = M M 0 V R T \pi=CRT\Rightarrow \pi=\frac{n}{V}RT\Rightarrow \pi=\frac{M}{M_{0}V}RT Substituting values with some approximations, M 0 = 11500 g m o l − 1 M_{0}=11500 g\ mol^{-1}

Symbols Used:

  1. π \pi =Osmotic Pressure
  2. C =Concentration (Molarity)
  3. R =Gas constant (0.0821 L a t m K − 1 L\ atmK^{-1} )
  4. T =Temperature
  5. n =moles
  6. V =Volume
  7. M =Given mass
  8. M 0 M_{0} =Molar Mass
3 Helpful 0 Interesting 0 Brilliant 0 Confused

Not writing down the units to save time. 

(0.042)=(298)(0.08206)(((2.00)/(MM))/(0.100))

from where we get 

MM = 11644.7
1 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Feb 16, 2016

http://www.chemteam.info/Solutions/WS-Osmosis-AP.html

π = i M R T \pi = i~M~R~T where i ≈ 1 i \approx 1

0.042 ≈ \approx 1 × \times M × \times 0.08206 × \times 298

M ≈ \approx 1.7175 × 1 0 − 3 m o l / L ≈ 2 g 0.1 L ÷ 11500 g / m o l ≈ 1.7391 × 1 0 − 3 m o l / L \times 10^{-3} mol/ L \approx \frac{2 g}{0.1 L} \div 11500 g/ mol \approx 1.7391 \times 10^{-3} mol/ L

i = 1.7175 1.7391 ≈ 0.9876 \frac{1.7175}{1.7391} \approx 0.9876

Note: 2 g 0.1 L \frac{2 g}{0.1 L} is also an approximation.

Answer: 11 , 500 g m o l − 1 \boxed{11,500~g~mol^{-1}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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