Answer
Verified
392.4k+ views
Hint: We know that Henry's law is used to determine the unknown pressure quantity and unknown mole fraction at respective temperatures.
Complete step by step answer:
Henry's constant for oxygen dissolved in water is $4.34 \times {10^4}\;{\rm{atm}}$.
The temperature is ${\rm{25^\circ C}}$.
The partial pressure of oxygen in air is ${\rm{0}}{\rm{.2}}\;{\rm{atm}}$.
The henry law can be defined by with the help of vapour pressure and number of mole fraction of gas in respective solutions. The mathematical form of henry law is depicted below.
${\rm{P}} = {{\rm{K}}_{\rm{H}}} \times {\rm{X}}$
Where, ${{\rm{K}}_{\rm{H}}}$ is the henry constant, P is the vapour pressure, and X is the mole fraction.
The mole fraction of oxygen can be calculated with the help of henry haw.
${{\rm{X}}_{{{\rm{o}}_{\rm{2}}}}} = \dfrac{{\rm{P}}}{{{{\rm{K}}_{\rm{H}}}}}$
Substitute the value of henry constant and pressure in the above equation.
$\begin{array}{c}
{{\rm{X}}_{{{\rm{o}}_{\rm{2}}}}} = \dfrac{{0.2\;{\rm{atm}}}}{{4.34 \times {{10}^4}\;{\rm{atm}}}}\\
= 4.6 \times {10^{ - 6}}
\end{array}$
That means $4.6 \times {10^{ - 6}}$ moles of oxygen gas present in one mole of water.
And $4.6 \times {10^{ - 6}}$ moles oxygen in ${\rm{18}}\;{\rm{g}}$ of water.
$\left( {\dfrac{{18}}{{18}} = 1\;{\rm{mole}}} \right)\left( {{\rm{18}}\;{\rm{g/mL}}} \right)$
The density of water is ${\rm{1}}\;{\rm{g/mL}}$.
So, it means ${\rm{18}}\;{\rm{g}} = {\rm{18}}\;{\rm{mL}}$
The conversion of milliliters to liters is done as follows.
$\begin{array}{c}
{\rm{1}}\;{\rm{mL}} = 0.001\;{\rm{L}}\\
{\rm{18}}\;{\rm{mL}} = 18 \times 0.001\;{\rm{L}}\\
= 18 \times {10^{ - 3}}\;{\rm{L}}
\end{array}$
Now, calculate the concentration by using the formula given below.
${\rm{Concentration}} = \dfrac{{{\rm{Moles}}}}{{{\rm{Volume}}}}$
Substitute the respective values of moles and volume in the above equation. We get,
\[\begin{array}{c}
{\rm{Concentration}} = \dfrac{{4.6 \times {{10}^{ - 6}}\;{\rm{mol}}}}{{18 \times {{10}^{ - 3}}\;{\rm{L}}}}\\
= 2.55 \times {10^{ - 4}}\;{\rm{M}}
\end{array}\]
Therefore, the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at ${\rm{25^\circ C}}$ is \[2.55 \times {10^{ - 4}}\;{\rm{M}}\].
Note:
The concentration or molarity of the aqueous solution usually be affected by the increasing or decreasing of temperature. The unit representation of the molarity is mol/L.
Complete step by step answer:
Henry's constant for oxygen dissolved in water is $4.34 \times {10^4}\;{\rm{atm}}$.
The temperature is ${\rm{25^\circ C}}$.
The partial pressure of oxygen in air is ${\rm{0}}{\rm{.2}}\;{\rm{atm}}$.
The henry law can be defined by with the help of vapour pressure and number of mole fraction of gas in respective solutions. The mathematical form of henry law is depicted below.
${\rm{P}} = {{\rm{K}}_{\rm{H}}} \times {\rm{X}}$
Where, ${{\rm{K}}_{\rm{H}}}$ is the henry constant, P is the vapour pressure, and X is the mole fraction.
The mole fraction of oxygen can be calculated with the help of henry haw.
${{\rm{X}}_{{{\rm{o}}_{\rm{2}}}}} = \dfrac{{\rm{P}}}{{{{\rm{K}}_{\rm{H}}}}}$
Substitute the value of henry constant and pressure in the above equation.
$\begin{array}{c}
{{\rm{X}}_{{{\rm{o}}_{\rm{2}}}}} = \dfrac{{0.2\;{\rm{atm}}}}{{4.34 \times {{10}^4}\;{\rm{atm}}}}\\
= 4.6 \times {10^{ - 6}}
\end{array}$
That means $4.6 \times {10^{ - 6}}$ moles of oxygen gas present in one mole of water.
And $4.6 \times {10^{ - 6}}$ moles oxygen in ${\rm{18}}\;{\rm{g}}$ of water.
$\left( {\dfrac{{18}}{{18}} = 1\;{\rm{mole}}} \right)\left( {{\rm{18}}\;{\rm{g/mL}}} \right)$
The density of water is ${\rm{1}}\;{\rm{g/mL}}$.
So, it means ${\rm{18}}\;{\rm{g}} = {\rm{18}}\;{\rm{mL}}$
The conversion of milliliters to liters is done as follows.
$\begin{array}{c}
{\rm{1}}\;{\rm{mL}} = 0.001\;{\rm{L}}\\
{\rm{18}}\;{\rm{mL}} = 18 \times 0.001\;{\rm{L}}\\
= 18 \times {10^{ - 3}}\;{\rm{L}}
\end{array}$
Now, calculate the concentration by using the formula given below.
${\rm{Concentration}} = \dfrac{{{\rm{Moles}}}}{{{\rm{Volume}}}}$
Substitute the respective values of moles and volume in the above equation. We get,
\[\begin{array}{c}
{\rm{Concentration}} = \dfrac{{4.6 \times {{10}^{ - 6}}\;{\rm{mol}}}}{{18 \times {{10}^{ - 3}}\;{\rm{L}}}}\\
= 2.55 \times {10^{ - 4}}\;{\rm{M}}
\end{array}\]
Therefore, the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at ${\rm{25^\circ C}}$ is \[2.55 \times {10^{ - 4}}\;{\rm{M}}\].
Note:
The concentration or molarity of the aqueous solution usually be affected by the increasing or decreasing of temperature. The unit representation of the molarity is mol/L.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE