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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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Calculate osmotic pressure of a solutions having $0.1\; M\;NaCl$ and $0.2\; M\; Na_2SO_4$ and $0.5 M\; HA$. (Given: Weak acid is 20% dissociated at $300^{\circ}K$)

$\begin{array}{1 1} 34.482\;atm \\ 16.872\;atm \\ 24.694\;atm \\ 36.762\;atm \end{array}$
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Answer: 34.482 atm
The external pressure which must be applied on solution side to stop the process of osmosis is called osmotic pressure of the solution. It is calculated as $\pi = \Sigma \;i_nC_nRT$
In this case $\pi_{NaCl} = 0.1\; \times 2 RT$, $\pi_{Na_2SO_4} = 0.2 \times 3 RT$ and $\pi_{HA} = 0.5\; RT \times 1.2$
$R$ is the ideal gas constant $ = 0.082 \; $$\text{lit.atm.mol}^{-1}$$ \text{K}^{-1}$
$\Rightarrow \pi = 0.0821 \times 300 (0.2 + 0.6 + 0.6) = 34.482\; atm$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai
 

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