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Questions  >>  CBSE XII  >>  Chemistry  >>  Solutions

The Henry's law constant for oxygen dissolved in water is $4.34\times 10^4$ atm at $25^{\large\circ}$C.If the partial pressure of oxygen in air is 0.2atm,under atmospheric conditions,calculate the concentration (in moles per litre) of dissolve oxygen in water in equilibrium with air at $25^{\large\circ}$C.

$\begin{array}{1 1}2.55\times 10^{-4}M\\3.55\times 10^{-4}M\\4.55\times 10^{-4}M\\1.55\times 10^{-4}M\end{array} $

1 Answer

According to Henry's law
Given :
$P=K_H x$
$K_H=4.34\times 10^4$ atm
$P_{O_2}=0.2$ atm
$P_{O_2}=K_H x_{O_2}$
$\therefore x_{O_2}=\large\frac{P_{O_2}}{K_H}$
$\Rightarrow \large\frac{0.2}{4.34\times 10^4}$
$\Rightarrow 4.6\times 10^{-6}$
Changing mole fraction into molarity
Moles of water =$\large\frac{1000}{18}$
$\Rightarrow 55.5$ mol
Now $n_{O_2}=\large\frac{n_{O_2}}{n_{H_2O}+n_{O_2}}$
But $n_{O_2}$ is very small in comparision to $n_{H_2O}$
$n_{O_2}+n_{H_2O}\approx n_{H_2O}$
$\therefore x_{O_2}=\large\frac{n_{O_2}}{n_{H_2O}}$
$n_{O_2}=4.6\times 10^{-6}\times 55.5$
$\Rightarrow 2.55\times 10^{-4}$ mol
Since $2.55\times 10^{-4}$ mol are present in 1000ml of solution,
Molarity =$2.55\times 10^{-4}M$
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