Determine the osmotic pressure of a solution prepared by

Question:

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.

Solution:

When $\mathrm{K}_{2} \mathrm{SO}_{4}$ is dissolved in water, $\mathrm{K}^{+}$and $\mathrm{SO}_{4}^{2-}$ ions are produced.

$\mathrm{K}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-}$

Total number of ions produced = 3

=3

Given,

w = 25 mg = 0.025 g

V = 2 L

T = 250C = (25 + 273) K = 298 K

Also, we know that:

R = 0.0821 L atm K-1mol-1

M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol-1

Appling the following relation,

$\pi=i \frac{n}{v} \mathrm{R} T$

$=i \frac{w}{M} \frac{1}{v} \mathrm{R} T$

$=3 \times \frac{0.025}{174} \times \frac{1}{2} \times 0.0821 \times 298$

$=5.27 \times 10^{-3} \mathrm{~atm}$

 

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