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Karnataka Board PUCPUC Science 2nd PUC Class 12

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

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

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 liter of water at 25°C, assuming that it is completely dissociated. (mol. wt. of K2SO4 = 174 g mol−1)

Numerical
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Solution

Given: K2SO4 dissolved (w) = 25 mg = 0.025 g

Volume of solution (V) = 2 L

Temperature (T) = 25°C = (25 + 273) K = 298 K

R = 0.0821 L atm K−1 mol−1

When K2SO4 is dissolved in water, K+ and \[\ce{SO^{2-}4}\] ions are produced.

\[\ce{K2SO4 -> 2K+ + SO^{2-}4}\]

Total number of ions produced = 3

∴ i = 3

Molar mass of K2SO4 (M) = (2 × 39) + (1 × 32) + (4 × 16)

= 174 g mol−1

Applying the following relation,

π = `i n/V RT`

= `i xx w/M xx 1/V RT`

= `3 xx 0.025/174 xx 1/2 xx 0.0821 xx 298`

= 5.27 × 10−3 atm

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Chapter 1: Solutions - 'NCERT TEXT-BOOK' Exercises [Page 129]

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Nootan Chemistry [English] Class 12 ISC
Chapter 1 Solutions
'NCERT TEXT-BOOK' Exercises | Q 2.41 | Page 129
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NCERT Chemistry Part 1 and 2 [English] Class 12
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Exercises | Q 1.41 | Page 30

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