Respuesta :
let the solubility be x.
solubility is directly proportional to the pressure.so,
4.04*10^-4/x=1/0.169 (since 760 mm = 1 atm)
or x=6.827*10^-5 g/100 mL
The solubility of [tex]C{O_2}[/tex]in an open bottle of seltzer water is[tex]\boxed{{\mathbf{9}}.{\mathbf{3394g}}{\text{/}}{\mathbf{100}}{\text{ }}{\mathbf{mL}}}[/tex].
Further explanation
Molar solubility is the concentration of pure substance dissolved in saturated solution. It is generally defined as ratio of number of moles of substance dissolved in one litre of solution until the solution becomes saturated. The S.I unit of molar solubility is mol/L.
The equilibrium constant between the compound and its ion, when dissolved in solution, is known as solubility product constant. It is denoted by [tex]{{\text{K}}_{{\text{sp}}}}[/tex]. The solubility product constant is used to calculate the product of concentration of ions at equilibrium. Higher the solubility product constant more will be the solubility of the compound.
The formula to relate concentration (C), volume (V), and number of moles (n) is as follows:
[tex]{\text{C = }}\frac{{\text{n}}}{{\text{V}}}[/tex] …… (1)
Ideal gas law is considered as the equation of state for any hypothetical gas. The expression for the ideal gas equation of gas is as follows:
[tex]{\text{PV}}={\text{nRT}}[/tex] …… (2)
Here,
P is the pressure of the gas.
V is the volume of gas.
n denotes the number of moles of gas.
R is the gas constant.
T is the temperature of gas.
The equation (2) can be written as:
[tex]{\text{P}}=\frac{{\text{n}}}{{\text{V}}}{\text{RT}}[/tex] …… (3)
Substitute C for [tex]\frac{{\text{n}}}{{\text{V}}}[/tex] in equation (2), the formula will become,
[tex]{\text{P}} = {\text{CRT}}[/tex] …… (4)
If the temperature is constant then RT will be constant. Then the equation (4) can be written as follows:
[tex]\frac{{\text{P}}}{{\text{C}}}={\text{constant}}[/tex] …… (5)
If we have initial and final value of pressure and concentration then the equation (5) will be written as follows:
[tex]{{\text{P}}_1}{{\text{C}}_{\text{2}}}={{\text{P}}_2}{{\text{C}}_1}[/tex] …… (6)
Here,
[tex]{{\text{P}}_1}[/tex]is the initial pressure.
[tex]{{\text{P}}_2}[/tex]is the final pressure.
[tex]{{\text{C}}_1}[/tex]is the initial concentration.
[tex]{{\text{C}}_2}[/tex]is the final concentration.
On rearranging equation (6) for [tex]{{\text{C}}_2}[/tex] we get,
[tex]{{\text{C}}_{\text{2}}}=\frac{{{{\text{P}}_{\text{2}}}{{\text{C}}_{\text{1}}}}}{{{{\text{P}}_{\text{1}}}}}[/tex] …… (7)
[tex]{{\text{P}}_1}[/tex]is 760 mmHg
[tex]{{\text{P}}_2}[/tex]is[tex]{\text{4}}{\text{.2}} \times {\text{1}}{{\text{0}}^{\text{4}}}\;{\text{mmHg}}[/tex]
[tex]{{\text{C}}_1}[/tex]is 0.169g/100 mL.
Substitute the values of [tex]{{\text{P}}_1}[/tex], [tex]{{\text{P}}_2}[/tex], and [tex]{{\text{C}}_1}[/tex]in equation (7).
[tex]\begin{aligned}{{\text{C}}_2}&=\frac{{\left({{\text{4}}{\text{.2}} \times {\text{1}}{{\text{0}}^{\text{4}}}\;{\text{mmHg}}}\right)\times\left( {{\text{0}}{\text{.169 g/100 mL}}}\right)}}{{\left({{\text{760 mmHg}}}\right)}}\\&={\text{9}}{\text{.3394 g/100}}\;{\text{mL}}\\\end{aligned}[/tex]
The solubility of [tex]{\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}[/tex]in an open bottle of seltzer water is 9.3394g/100 mL.
Learn more:
1. Determination of number of atoms of hydrogen are present in 2.92g of water molecule https://brainly.com/question/899408
2. Identify the intermolecular forces present between given molecules. https://brainly.com/question/10107765
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Solutions
Keywords: ideal gas, pressure, volume, temperature, number of moles, initial, final, equation, 760 mmHg, 9.334g/100mL, 0.169g/100mL.
