Question

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C assuming that it is completely dissociated.​

Answer 1

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K2SO4 dissolved = 25mg = 0.025 g

Volume of solution = 2 L

T = 25°C = 298 K

Molar mass of K2SO4 = 2 × 39 + 32 + 4 × 16 = 174 g mol-¹

As K2SO4 dissociates completely as K2SO4 ----) 2K+ + SO4²-

i.e., ions produced = 3

Therefore, i = 3

again,

$\pi =$

iCRT = i × n/v RT = i × w/M × 1/V RT =

3 × 0.025g/174g mol-¹ × 1/2L × 0.0821 L atm K-¹ mol-¹ × 298 K

= 5.27 × 10-³ atm.

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Answer 2

Answer:

Given that: Mass of K2SO4

w=25mg=0.025g

Volume V=2L

T=25+273=298K

The reaction of dissociation of K2SO4

:K2SO4→2K+ +SO42−

Number if ions produced = 2+ 1 = 3

So, van’t Hoff factor

i=3

Use the formula of Osmotic pressure as:

π=n/VRT

π=wRT/MV

where R= Gas constant = 0.0821L atm K−1mol−1

Molar mass of K2SO4=M=2×39+1×32+4×16=174gmol

Upon substitution we get:

π=(3× 0.025×0.0821×298)/174×2

π=5.27×10−3 atm