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Solutions

Question
CBSEENCH12010160

Determine the osmotic pressure of a solution prepared by dissolving 2.5 x 10-2 g of K2SO4 in 2L of water at 250C, assuming that it is completely dissociated. 

(R = 0.0821 L atm K-1   mol-1, Molar mass of K2SO4 = 174 g mol-1)

Solution

w2 = 2.5 x 10-2 g (Mass of K2SO4)

M2 = 174 g mol-1 (Molar mass of K2SO4)

V = 2L,

R = 0.0821 L atm K-1 mol-1 and

T = 25°C = 298 K

 Osmotic pressure,  
  straight pi space equals fraction numerator straight w subscript 2 space straight x space straight R space straight x space straight T over denominator straight M subscript 2 space straight x space straight V end fraction

straight pi space fraction numerator 2.5 space straight x space 10 to the power of negative 2 end exponent space straight x space 0.0821 space straight x space 298 over denominator 174 space space straight x space 2 end fraction space equals fraction numerator space 61.1645 space straight x space 10 to the power of negative 2 end exponent over denominator 348 end fraction space equals space 1.76 space straight x space 10 to the power of negative 3 end exponent space atm