Answer
414.9k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)