Solubility Questions in English

(C) According to Henry's law,the solubility $(S)$ of a gas in a liquid is given by the formula: $S = K_H \times P$
Where:
$K_H = 6.8 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$
$P = 0.8 \ atm$
Substituting the values:
$S = (6.8 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}) \times (0.8 \ atm)$
$S = 5.44 \times 10^{-4} \ mol \ dm^{-3}$
Since $1 \ mol \ dm^{-3} = 1 \ M$,the solubility is $5.44 \times 10^{-4} \ M$.

(A) According to Henry's Law,the solubility $(S)$ of a gas is given by the formula: $S = K_{H} \times P$.
Given: $K_{H} = 0.16 \ mol \ dm^{-3} \ bar^{-1}$ and $P = 0.15 \ bar$.
Substituting the values: $S = 0.16 \ mol \ dm^{-3} \ bar^{-1} \times 0.15 \ bar = 0.024 \ mol \ dm^{-3}$.
Therefore,$S = 2.4 \times 10^{-2} \ mol \ dm^{-3}$.

(A) According to Henry's law,the solubility $(S)$ is related to the partial pressure $(P)$ by the equation: $S = K_H P$
Here,$K_H$ is the Henry's law constant.
Rearranging the formula to solve for $K_H$: $K_H = \frac{S}{P}$
Substituting the given values: $K_H = \frac{5.14 \times 10^{-4} \ mol \ dm^{-3}}{0.75 \ bar} = 6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$

(A) According to Henry's Law,the solubility $(S)$ of a gas is given by the formula: $S = K_{H} \times P$.
Given: $K_{H} = 0.16 \ mol \ dm^{-3} \ atm^{-1}$ and $P = 0.18 \ atm$.
Substituting the values: $S = 0.16 \times 0.18 = 0.0288 \ mol \ dm^{-3}$.
Rounding to two significant figures,we get $0.029 \ mol \ dm^{-3}$.

(A) According to Henry's law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid surface.
Mathematically,$P = K_H \cdot x$,where $P$ is the partial pressure,$x$ is the mole fraction (solubility),and $K_H$ is Henry's law constant.

(A) According to Henry's law,the solubility $(S)$ of a gas is given by the formula:
$S = K_{H} \times P$
Given:
$K_{H} = 0.159 \ mol \ dm^{-3} \ bar^{-1}$
$P = 0.346 \ bar$
Calculation:
$S = 0.159 \ mol \ dm^{-3} \ bar^{-1} \times 0.346 \ bar = 0.055 \ mol \ dm^{-3}$
Thus,the solubility of the gas is $0.055 \ mol \ dm^{-3}$.

(A) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to its partial pressure $(P)$: $S = K_{H} \times P$.
Here,$S = 0.028 \ mol \ dm^{-3}$ and $P = 0.346 \ bar$.
Therefore,the Henry's law constant $(K_{H})$ is calculated as:
$K_{H} = \frac{S}{P} = \frac{0.028 \ mol \ dm^{-3}}{0.346 \ bar} \approx 0.081 \ mol \ dm^{-3} \ bar^{-1}$.

(D) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to the partial pressure $(P)$ of the gas above the solution:
$S = K_H \times P$
Given:
Solubility $(S)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3}$
Henry's law constant $(K_H)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$
Rearranging the formula to solve for pressure $(P)$:
$P = S / K_H$
$P = (6.85 \times 10^{-4} \ mol \ dm^{-3}) / (6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1})$
$P = 1.0 \ bar$

(B) According to Henry's law,the solubility of a gas is given by the formula: $S = k \times p$
Here,$k = 0.45 \text{ mol dm}^{-3} \text{ atm}^{-1}$ and $p = 1.2 \text{ atm}$.
Substituting the values: $S = 0.45 \text{ mol dm}^{-3} \text{ atm}^{-1} \times 1.2 \text{ atm} = 0.54 \text{ mol dm}^{-3}$.
Thus,the solubility of the gas is $0.54 \text{ mol dm}^{-3}$.

(D) Henry's law states that the partial pressure of the gas in the vapor phase $(p)$ is proportional to the mole fraction of the gas $(x)$ in the solution.
Mathematically,it is expressed as $p = K_H \cdot x$,where $K_H$ is Henry's law constant.
This law directly describes the quantitative relationship between the solubility of a gas in a liquid and its pressure.

(C) According to Henry's law,the solubility $(S)$ of a gas in a liquid is directly proportional to the partial pressure $(P)$ of the gas above the liquid: $S = K_H \times P$.
Here,$S$ is expressed in $mol \ dm^{-3}$ and $P$ is expressed in $bar$.
Therefore,the unit of Henry's law constant $(K_H)$ is given by: $K_H = \frac{S}{P} = \frac{mol \ dm^{-3}}{bar} = mol \ dm^{-3} \ bar^{-1}$.

(A) According to Henry's law,the solubility $(S)$ of a gas is given by the formula: $S = K_H \times P$
Given:
Pressure $(P)$ = $0.8 \ atm$
Henry's law constant $(K_H)$ = $6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$
Substituting the values:
$S = 6.85 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1} \times 0.8 \ atm$
$S = 5.48 \times 10^{-4} \ mol \ dm^{-3}$

(B) According to Henry's law,the solubility $(S)$ of a gas is given by $S = K_H \times p$.
First,convert the pressure from $mm \ Hg$ to $atm$:
$p = \frac{260 \ mm \ Hg}{760 \ mm \ Hg \ atm^{-1}} \approx 0.3421 \ atm$.
Now,calculate the solubility:
$S = 0.159 \ mol \ dm^{-3} \ atm^{-1} \times 0.3421 \ atm \approx 0.05439 \ mol \ dm^{-3}$.
Rounding to two significant figures,we get $S = 5.4 \times 10^{-2} \ mol \ dm^{-3}$.

(C) According to Henry's Law,the solubility of a gas is given by the formula: $\text{Solubility} = K_H \times P_{gas}$.
Given,$K_H = 0.16 \ mol \ L^{-1} \ bar^{-1}$ and $\text{Solubility} = 0.08 \ mol \ L^{-1}$.
Rearranging the formula to solve for pressure: $P_{gas} = \frac{\text{Solubility}}{K_H}$.
Substituting the values: $P_{gas} = \frac{0.08 \ mol \ L^{-1}}{0.16 \ mol \ L^{-1} \ bar^{-1}} = 0.5 \ bar$.

(B) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to the partial pressure $(p)$ of the gas: $S = K_H \times p$.
Given:
Solubility $(S)$ = $7 \times 10^{-4} \ mol \ L^{-1}$
Pressure $(p)$ = $1 \ bar$
Substituting the values into the formula:
$7 \times 10^{-4} \ mol \ L^{-1} = K_H \times 1 \ bar$
Therefore,$K_H = \frac{7 \times 10^{-4} \ mol \ L^{-1}}{1 \ bar} = 7 \times 10^{-4} \ mol \ L^{-1} \ bar^{-1}$.

(C) According to Henry's law,$p = K_H \times x$,where $p$ is the partial pressure of the gas,$K_H$ is the Henry's law constant,and $x$ is the mole fraction of the gas in the solution.
$1$. $K_H$ depends on the nature of the gas and increases with an increase in temperature.
$2$. From the relation $p = K_H \times x$,we can write $x = p / K_H$. This shows that for a given pressure $p$,the solubility $(x)$ is inversely proportional to $K_H$.
$3$. Therefore,a higher value of $K_H$ implies lower solubility of the gas in the liquid.
$4$. Thus,the statement 'Higher the value of $K_H$ at a given pressure,higher is the solubility of the gas in the liquids' is incorrect.

(A) According to Henry's law,the solubility $(S)$ of a gas is given by the formula: $S = K_{H} \times P$
Given: $K_{H} = 1.3 \times 10^{-3} \ mol \ dm^{-3} \ atm^{-1}$ and $P = 0.46 \ atm$
Substituting the values: $S = (1.3 \times 10^{-3} \ mol \ dm^{-3} \ atm^{-1}) \times (0.46 \ atm) = 5.98 \times 10^{-4} \ mol \ dm^{-3}$

(C) Henry's law is given by the relation $P = K_H \cdot x$,where $P$ is the partial pressure of the gas,$K_H$ is Henry's law constant,and $x$ is the mole fraction of the gas in the solution.
For a given partial pressure,the solubility of a gas $(x)$ is inversely proportional to $K_H$ $(x = P / K_H)$.
As the temperature increases,the kinetic energy of gas molecules increases,making it harder for them to remain dissolved in the solvent,thus decreasing solubility.
Since solubility decreases as temperature increases,the value of $K_H$ must increase with an increase in temperature for a given pressure.

(A) According to Henry's Law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution.
Mathematically,this is expressed as $S = K_{H} \times P$.
Here,$S$ represents the solubility of the gas,$P$ represents the partial pressure of the gas,and $K_{H}$ is the Henry's law constant.

(D) Henry's law is expressed as $P = K_H \times C$,where $P$ is the partial pressure of the gas and $C$ is the concentration of the gas in the solution.
Rearranging for the constant,we get $K_H = \frac{P}{C}$.
The unit of pressure $P$ is $atm$ and the unit of concentration $C$ is $mol \ dm^{-3}$.
Therefore,the unit of $K_H$ is $\frac{atm}{mol \ dm^{-3}} = atm \ dm^3 \ mol^{-1}$ or $mol^{-1} \ dm^3 \ atm$. However,in many textbooks,$K_H$ is expressed as $P = K_H \times x$ (where $x$ is mole fraction,unitless),making the unit of $K_H$ simply $atm$. If $K_H$ is defined as $C = K_H \times P$,then the unit is $mol \ dm^{-3} \ atm^{-1}$. Given the options,the correct unit for $K_H$ in the context of solubility concentration is $mol \ dm^{-3} \ atm^{-1}$.

(D) Henry's law states that the mass of a gas dissolved in a given volume of a liquid at a constant temperature is directly proportional to the pressure of the gas in equilibrium with the liquid.
Thus,it defines the relationship between the solubility of a gas in a liquid and the external pressure at a constant temperature.

(C) Henry's law states that the solubility of a gas in a liquid is proportional to the partial pressure of the gas above the liquid,given by the equation $P = K_H \cdot x$,where $K_H$ is Henry's constant.
As the temperature increases,the kinetic energy of the gas molecules increases,making it easier for them to escape from the solution.
Consequently,the solubility of the gas decreases.
Since $K_H = P / x$,for a constant partial pressure $P$,an increase in temperature leads to a decrease in the mole fraction $x$ of the dissolved gas.
Therefore,the value of $K_H$ increases with an increase in temperature.

(D) The solubility of a solid solute in a liquid solvent depends on the nature of the solute and solvent (like dissolves like) and the temperature.
However,solids are incompressible,so the effect of pressure on the solubility of a solid in a liquid is negligible.
Therefore,the maximum amount of solid solute that can be dissolved does not depend upon pressure $(iii)$.
Thus,the correct option is $D$.

(D) The solubility of gases in liquids generally decreases with an increase in temperature because the dissolution of gas in a liquid is an exothermic process. According to Le Chatelier's principle,an increase in temperature shifts the equilibrium in the backward direction,thereby decreasing the solubility. Among the given options,$Chlorine$ is a gas,so its solubility in water decreases as the temperature increases.

(C) According to Henry's law,the partial pressure of a gas $(p)$ is directly proportional to its mass $(m)$ dissolved in a given volume of solvent at a constant temperature: $p \propto m$ or $p = k \times m$.
Given: $p_1 = 1 \ bar$,$m_1 = 7.14 \times 10^{-3} \ g$.
We need to find $p_2$ when $m_2 = 5 \times 10^{-2} \ g$.
Using the ratio: $\frac{p_2}{p_1} = \frac{m_2}{m_1}$.
$\frac{p_2}{1} = \frac{5 \times 10^{-2}}{7.14 \times 10^{-3}} = \frac{50 \times 10^{-3}}{7.14 \times 10^{-3}} \approx 7 \ bar$.
Thus,the partial pressure is $7 \ bar$.

(A) solid solution is a solution in which the solute and solvent are both in the solid state.
$A$ Copper dissolved in gold is an example of a solid solution (alloy),where both components are solids.
$B$ Glucose dissolved in water is a solid-in-liquid solution.
$C$ Camphor in nitrogen gas is a solid-in-gas solution.
$D$ Ethanol dissolved in water is a liquid-in-liquid solution.
Therefore,the correct answer is $A$.

(D) solution where the solute is a gas and the solvent is a solid is known as a gas-in-solid solution.
In the case of $H_2$ gas adsorbed on a palladium $(Pd)$ metal surface,hydrogen acts as the solute (gas) and palladium acts as the solvent (solid).
Therefore,the correct option is $D$.

(B) According to Henry's Law,the solubility of a gas in a liquid is given by the relation $P = K_H \times x$,where $P$ is the partial pressure of the gas,$K_H$ is Henry's law constant,and $x$ is the mole fraction of the gas in the solution.
Rearranging for solubility $(x)$,we get $x = P / K_H$.
Since $K_H$ is in the denominator,as the temperature increases,$K_H$ increases,which leads to a decrease in the solubility $(x)$ of the gaseous solute.

(A) solution where the solute is liquid and the solvent is solid is known as a solid solution of a liquid in a solid. $Zinc$ amalgam is an example of this,where liquid mercury $(Hg)$ acts as the solute and solid $Zinc$ $(Zn)$ acts as the solvent.

(B) Moles of $N_2 = \frac{5.5 \times 10^{-3} \text{ g}}{28 \text{ g/mol}} \approx 1.96 \times 10^{-4} \text{ mol}$.
Moles of $H_2O = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ mol}$.
Mole fraction of $N_2$ at $1 \text{ atm}$ $(x_1)$ = $\frac{n_{N_2}}{n_{H_2O} + n_{N_2}} \approx \frac{1.96 \times 10^{-4}}{10} = 1.96 \times 10^{-5}$.
According to Henry's law,$p = K_H \cdot x$,which implies $x \propto p$.
At $5 \text{ atm}$ pressure,the new mole fraction $(x_2)$ is $x_2 = 5 \times x_1 = 5 \times 1.96 \times 10^{-5} = 9.8 \times 10^{-5} \approx 1 \times 10^{-4}$.

(D) The dissolution of a gas in a liquid is an exothermic process. According to Le Chatelier's principle,for an exothermic process,an increase in temperature shifts the equilibrium in the backward direction,thereby decreasing the solubility of the gas in the liquid.
Therefore,the solubility of a gas in a liquid decreases as the temperature increases.
This is correctly represented by the graph where solubility decreases linearly or non-linearly with an increase in temperature,which matches option $D$.

(C) According to Henry's law: $P = K_{H} X$
First,calculate the mole fraction of $N_2$ $(X_{N_2})$:
$X_{N_2} = \frac{P_{N_2}}{K_{H}} = \frac{0.987 \ bar}{76.48 \times 10^3 \ bar} = 1.29 \times 10^{-5}$
Calculate the number of moles of water in $1 \ litre$ $(1000 \ g)$:
$n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} = 55.5 \ mol$
Since $n_{N_2} \ll n_{H_2O}$,we use the approximation $X_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}}$:
$n_{N_2} = X_{N_2} \times n_{H_2O} = 1.29 \times 10^{-5} \times 55.5 \ mol = 7.16 \times 10^{-4} \ mol$

(B) According to Henry's Law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas $(p)$ above the liquid surface.
Additionally,the dissolution of a gas in a liquid is an exothermic process. According to Le Chatelier's principle,decreasing the temperature $(T)$ favors the forward reaction,thereby increasing the solubility of the gas.

(A) Given,Henry's law constant $(K_H)$ for the solubility of $N_2$ gas in water at $298 \ K = 1 \times 10^5 \ atm$.
Mole fraction of $N_2$ in air $(\chi_{N_2, \text{air}}) = 0.8$.
Total pressure $(P_{\text{total}}) = 5 \ atm$.
Partial pressure of nitrogen $(p_{N_2}) = P_{\text{total}} \times \chi_{N_2, \text{air}} = 5 \times 0.8 = 4 \ atm$.
According to Henry's law,$p_{N_2} = K_H \times \chi_{N_2, \text{water}}$.
$4 = 10^5 \times \chi_{N_2, \text{water}} \Rightarrow \chi_{N_2, \text{water}} = 4 \times 10^{-5}$.
We know that $\chi_{N_2, \text{water}} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}}$.
Since $n_{N_2} \ll n_{H_2O}$,we can approximate $\chi_{N_2, \text{water}} \approx \frac{n_{N_2}}{n_{H_2O}}$.
$4 \times 10^{-5} = \frac{n_{N_2}}{10}$.
$n_{N_2} = 4 \times 10^{-4} \ moles$.

(C) Among the given statements,only statement $(c)$ is correct regarding the Henry's law constant $(K_{H})$.
$(a)$ $K_{H}$ is a function of the nature of the gas and the solvent used; therefore,the $K_{H}$ value is not the same for a gas in different solutions.
$(b)$ According to Henry's law,$p = K_{H} \cdot x$,which implies $x = p / K_{H}$. Thus,a higher $K_{H}$ value corresponds to lower solubility of the gas.
$(c)$ $K_{H}$ is temperature-dependent and generally increases with an increase in temperature,which explains why gases are less soluble in warm water.
$(d)$ Easily liquefiable gases have stronger intermolecular forces,which leads to lower $K_{H}$ values.

(C) According to Henry's law,the solubility of a gas in a liquid is inversely proportional to its Henry's law constant $(K_H)$.
Mathematically,$\text{Solubility} \propto \frac{1}{K_H}$.
Given $K_H$ values at $298 \ K$:
Argon $(I)$: $40.3 \ k\text{bar}$
Carbon dioxide $(II)$: $1.67 \ k\text{bar}$
Methane $(IV)$: $0.413 \ k\text{bar}$
Formaldehyde $(III)$: $1.83 \times 10^{-5} \ k\text{bar}$
Comparing the $K_H$ values: $1.83 \times 10^{-5} < 0.413 < 1.67 < 40.3$.
Therefore,the order of $K_H$ is $III < IV < II < I$.
Since solubility is inversely proportional to $K_H$,the increasing order of solubility is $I < II < IV < III$.

(C) The partition coefficient $(K)$ for silver between molten zinc and molten lead is $300$.
Let $x$ be the mass of $Ag$ in $Zn$ and $(1 - x)$ be the mass of $Ag$ in $Pb$.
Partition coefficient $= \frac{\text{Conc. of } Ag \text{ in molten } Zn}{\text{Conc. of } Ag \text{ in molten } Pb} = \frac{x / 10}{(1 - x) / 100} = 300$.
Solving for $x$: $\frac{10x}{1 - x} = 300 \implies 10x = 300 - 300x \implies 310x = 300 \implies x = \frac{30}{31}$.
Amount of silver in molten lead $= 1 - x = 1 - \frac{30}{31} = \frac{1}{31} \ g$.
Percentage of silver in lead $= \frac{1/31}{1} \times 100 \approx 3.22 \% \approx 3 \%$.

(A) According to Le Chatelier's principle,for an endothermic process,the dissolution can be represented as: $\text{Solute} + \text{Solvent} + \text{Heat} \rightleftharpoons \text{Solution}$.
When the temperature is increased,the equilibrium shifts in the forward direction to absorb the extra heat,thereby increasing the solubility of the solute.
The reason statement is also correct because a saturated solution represents a state of dynamic equilibrium where the rate of dissolution equals the rate of crystallization.
Since the shift in equilibrium due to temperature change is explained by Le Chatelier's principle,$(R)$ is the correct explanation for $(A)$.

(A) According to Henry's Law,the solubility of a gas in a liquid is inversely proportional to its Henry's Law constant $(K_{H})$.
Mathematically,$S \propto \frac{1}{K_{H}}$.
Given $K_{H}$ values are:
$H_2 = 71.18 \ kbar$
$CH_4 = 41.85 \ kbar$
$O_2 = 34.86 \ kbar$
$CO_2 = 1.67 \ kbar$
Since the solubility is inversely proportional to $K_{H}$,the gas with the highest $K_{H}$ value will have the lowest solubility.
The order of $K_{H}$ values is: $CO_2 < O_2 < CH_4 < H_2$.
Therefore,the increasing order of solubility is: $H_2 < CH_4 < O_2 < CO_2$.

(A) The gas bubbles (like $N_2$) in the bodily fluids affect nerve impulses,giving rise to decompression sickness or bends. These can be painful and fatal. To avoid bends sickness,the tanks used by scuba divers are filled with air diluted with helium gas $(He)$.

(B) According to Henry's law,$p = K_H \cdot \chi$,where $\chi$ is the mole fraction of $CO_2$.
Given $p = 3.4 \ bar$ and $K_H = 1.7 \times 10^3 \ bar$.
$\chi = \frac{p}{K_H} = \frac{3.4}{1.7 \times 10^3} = 2 \times 10^{-3}$.
Since the solution is dilute,$\chi = \frac{n_{CO_2}}{n_{H_2O}}$.
$n_{H_2O} = \frac{200 \ g}{18 \ g \ mol^{-1}} = 11.11 \ mol$.
$n_{CO_2} = \chi \times n_{H_2O} = 2 \times 10^{-3} \times 11.11 = 2.222 \times 10^{-2} \ mol$.
Molarity $= \frac{n_{CO_2}}{\text{Volume in } L} = \frac{2.222 \times 10^{-2} \ mol}{0.2 \ L} = 0.1111 \ mol \ L^{-1} = 1.11 \times 10^{-1} \ mol \ L^{-1}$.

(D) According to Henry's Law,$p = K_{H} \times x$.
Given: $p = 1 \ bar$,$K_{H} = 0.4 \ kbar = 400 \ bar$.
$x = \frac{p}{K_{H}} = \frac{1}{400} = 0.0025$.
Since $x = \frac{n_{CH_4}}{n_{CH_4} + n_{H_2O}} \approx \frac{n_{CH_4}}{n_{H_2O}}$,where $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
$n_{CH_4} = x \times n_{H_2O} = 0.0025 \times 55.55 = 0.1388 \ mol \approx 1.38 \times 10^{-1} \ mol$.

(C)

Gas	$K_H / \text{kbar}$ at $293 \text{ K}$
$He$	$144.97$
$H_2$	$69.16$
$O_2$	$34.86$
$N_2$	$76.48$

Based on the values provided for Henry's law constant $(K_H)$ at $293 \text{ K}$,$He$ has the highest value of $144.97 \text{ kbar}$.

(D) According to Henry's law: $P = x \cdot K_H$.
For $N_2$: $P_{N_2} = x_{N_2} \cdot K_H(N_2)$.
For $O_2$: $P_{O_2} = x_{O_2} \cdot K_H(O_2)$.
Given that $P_{N_2} = P_{O_2}$,$K_H(N_2) = 76.48 \ kbar$,and $K_H(O_2) = 34.86 \ kbar$.
The ratio of mole fractions is: $\frac{x_{N_2}}{x_{O_2}} = \frac{P_{N_2} / K_H(N_2)}{P_{O_2} / K_H(O_2)} = \frac{K_H(O_2)}{K_H(N_2)}$.
Substituting the values: $\frac{x_{N_2}}{x_{O_2}} = \frac{34.86}{76.48} \approx 0.45$.

(B) According to Henry's Law,$P_{O_2} = K_{H} \chi_{O_2}$.
Given: $P_{O_2} = 1 \ bar$,$K_{H} = 50 \ kbar = 50 \times 10^3 \ bar$.
$\chi_{O_2} = \frac{P_{O_2}}{K_{H}} = \frac{1}{50 \times 10^3} = 2 \times 10^{-5}$.
Since $\chi_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \frac{n_{O_2}}{n_{H_2O}}$,and $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} \approx 55.5 \ mol$.
$n_{O_2} = \chi_{O_2} \times 55.5 = 2 \times 10^{-5} \times 55.5 = 1.11 \times 10^{-3} \ mol$.
Mass of $O_2 = 1.11 \times 10^{-3} \ mol \times 32 \ g/mol = 0.03552 \ g$.
Concentration in $ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6 = \frac{0.03552 \ g}{1000 \ g} \times 10^6 = 35.52 \ ppm \approx 35.50 \ ppm$.

(D) According to Henry's law,the solubility of a gas in a liquid is inversely proportional to the Henry's law constant $(K_{H})$,given by the relation $S \propto \frac{1}{K_{H}}$.
Therefore,a lower $K_{H}$ value corresponds to higher solubility.
The given $K_{H}$ values are: $Ar = 40.39$,$CO_{2} = 1.67$,$CH_{4} = 0.413$,and $HCHO = 1.83 \times 10^{-5}$.
Comparing these values,the order of $K_{H}$ is: $Ar > CO_{2} > CH_{4} > HCHO$.
Thus,the increasing order of solubility is: $Ar < CO_{2} < CH_{4} < HCHO$.

(C) According to Henry's Law,$p = K_{H} \times \chi_{X}$,where $p$ is the partial pressure of the gas and $\chi_{X}$ is its mole fraction in the solution.
For the initial condition: $2 = K_{H} \times 0.02$ ...$(i)$
When the pressure is doubled,$p' = 4 \ bar$.
$4 = K_{H} \times \chi'_{X}$ ...$(ii)$
Dividing equation $(ii)$ by $(i)$: $\frac{4}{2} = \frac{\chi'_{X}}{0.02}$ $\Rightarrow 2 = \frac{\chi'_{X}}{0.02}$ $\Rightarrow \chi'_{X} = 0.04$.
The sum of mole fractions in a binary solution is $1$.
Therefore,the mole fraction of water is $\chi_{water} = 1 - \chi'_{X} = 1 - 0.04 = 0.96$.

(A) According to Henry's law,$p = K_H \times \chi$,where $p$ is the partial pressure,$K_H$ is Henry's law constant,and $\chi$ is the mole fraction of the gas.
Given: $p = 5 \ atm = 5 \times 1.01325 \times 10^5 \ Pa = 5.066 \times 10^5 \ Pa$,$K_H = 1.67 \times 10^8 \ Pa$.
Mole fraction $\chi = \frac{p}{K_H} = \frac{5.066 \times 10^5}{1.67 \times 10^8} \approx 3.033 \times 10^{-3}$.
Assuming $500 \ mL$ of water has a mass of $500 \ g$,the number of moles of water $n_{H_2O} = \frac{500}{18} \approx 27.78 \ mol$.
Since $\chi = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}}$,we have $n_{CO_2} = \chi \times n_{H_2O} = 3.033 \times 10^{-3} \times 27.78 \approx 0.08426 \ mol$.
Mass of $CO_2 = n_{CO_2} \times \text{molar mass} = 0.08426 \times 44 \approx 3.71 \ g$.

(A) Given: Pressure of $CO_2$ $(p)$ $= 3.34 \ bar = 3.34 \times 10^5 \ Pa$. Henry's law constant $(K_H)$ for $CO_2$ at $298 \ K$ is $1.67 \times 10^8 \ Pa$.
According to Henry's law,$p = K_H \times x$,where $x$ is the mole fraction of $CO_2$.
$x = \frac{p}{K_H} = \frac{3.34 \times 10^5 \ Pa}{1.67 \times 10^8 \ Pa} = 2 \times 10^{-3}$.
Volume of water $= 500 \ mL$. Since density of water is $1 \ g/mL$,mass of water $= 500 \ g$.
Moles of water $(n_{H_2O})$ $= \frac{500 \ g}{18 \ g/mol} = 27.78 \ mol$.
Since $x = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}}$,we have $n_{CO_2} = x \times n_{H_2O} = 2 \times 10^{-3} \times 27.78 \ mol = 0.05556 \ mol$.
Mass of $CO_2 = n_{CO_2} \times \text{Molar mass of } CO_2 = 0.05556 \ mol \times 44 \ g/mol = 2.4446 \ g \approx 2.442 \ g$.

(C) According to Henry's law: $p = K_H \times \chi$
Given: $p = 1.67 \ bar$,$K_H = 1.67 \ kbar = 1670 \ bar$.
Calculating mole fraction $\chi$ of $CO_2$: $\chi = \frac{p}{K_H} = \frac{1.67}{1670} = 0.001$.
Since the amount of $CO_2$ is very small,$\chi = \frac{n_{CO_2}}{n_{H_2O}} \approx 0.001$.
Number of moles of water in $1 \ L$ is $n_{H_2O} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$.
Therefore,$n_{CO_2} = 0.001 \times 55.55 = 0.05555 \ mol$.
Mass of $CO_2 = n_{CO_2} \times \text{Molar mass of } CO_2 = 0.05555 \ mol \times 44 \ g \ mol^{-1} \approx 2.44 \ g$.
Thus,the amount of $CO_2$ dissolved is $2.44 \ g \ L^{-1}$.