Solubility Questions in English

(D) According to Henry's law,$P = K_H \times \chi$,where $P$ is the partial pressure of the gas,$K_H$ is Henry's law constant,and $\chi$ is the mole fraction of the gas in the solution.
Given: $P = 1.67 \times 10^2 \ kPa = 1.67 \times 10^5 \ Pa$,$K_H = 1.67 \times 10^8 \ Pa$.
Calculating mole fraction $\chi = P / K_H = (1.67 \times 10^5) / (1.67 \times 10^8) = 10^{-3}$.
Since the solution is dilute,the mole fraction $\chi \approx n_{CO_2} / n_{H_2O}$.
For $1000 \ mL$ of water,mass $= 1000 \ g$,so $n_{H_2O} = 1000 / 18 = 55.55 \ mol$.
Thus,$n_{CO_2} = \chi \times n_{H_2O} = 10^{-3} \times 55.55 = 5.555 \times 10^{-2} \ mol$.
Since the volume is $1 \ L$,the concentration is $5.55 \times 10^{-2} \ mol \ L^{-1}$.

(C) According to Henry's law,the partial pressure of a gas is directly proportional to its mole fraction in the solution,given by $P = K_H \times \chi$.
Since the temperature remains constant,$K_H$ is constant.
When the pressure is doubled from $2 \ bar$ to $4 \ bar$,the mole fraction of the gas $(\chi_X)$ also doubles.
New mole fraction of gas $\chi_X' = 0.02 \times 2 = 0.04$.
The sum of mole fractions in a binary solution is $1$.
Therefore,the mole fraction of water $\chi_{H_2O} = 1 - \chi_X' = 1 - 0.04 = 0.96$.

(C) Henry's law states that the partial pressure of a gas in the vapor phase is proportional to the mole fraction of the gas in the solution. This law is valid only when the gas does not undergo any chemical reaction with the solvent.
$1$. Ammonia $(NH_3)$ reacts with water to form ammonium hydroxide $(NH_4OH)$.
$2$. Oxygen $(O_2)$ reacts with hemoglobin in the blood to form oxyhemoglobin.
$3$. Oxygen $(O_2)$ and Carbon dioxide $(CO_2)$ do not react chemically with water.
Therefore, Henry's law is valid for $C$ and $D$.

(C) Total pressure of air over water $= 1 \,atm$.
Partial pressure of $N_2$ and $O_2$ are:
$P_{N_2} = \frac{1 \times 79}{100} = 0.79 \,atm$
$P_{O_2} = \frac{1 \times 21}{100} = 0.21 \,atm$
Applying Henry's law $(P = K_H \times X)$:
$X_{N_2} = \frac{P_{N_2}}{K_{H, N_2}} = \frac{0.79}{8.57 \times 10^4} \approx 9.22 \times 10^{-6}$
$X_{O_2} = \frac{P_{O_2}}{K_{H, O_2}} = \frac{0.21}{4.56 \times 10^4} \approx 4.60 \times 10^{-6}$
Ratio of mole fractions of $N_2$ and $O_2$:
$\frac{X_{N_2}}{X_{O_2}} = \frac{9.22 \times 10^{-6}}{4.60 \times 10^{-6}} \approx 2$
Thus, the ratio is $2: 1$.

(C) According to Henry's law, $p = K_H \cdot x$, where $x$ is the mole fraction of the gas in the solution (solubility).
Therefore, solubility $x = \frac{p}{K_H}$.
At a constant pressure, solubility is inversely proportional to the Henry's law constant $(x \propto \frac{1}{K_H})$.
Given $K_H$ values: $HCHO \ (1.83 \times 10^{-5}) < CH_4 \ (0.413) < CO_2 \ (1.67) < Ar \ (40.3)$.
Thus, the order of solubility is: $HCHO > CH_4 > CO_2 > Ar$.

(B) At a total pressure of $5 \ atm$, the partial pressure of $N_2$ is $P_{N_2} = 5 \times 0.8 = 4 \ atm$.
According to Henry's Law, $P_{N_2} = K_{H} \times x_{N_2}$, where $x_{N_2}$ is the mole fraction of nitrogen gas dissolved in water.
Substituting the values: $4 \ atm = (1 \times 10^5 \ atm) \times x_{N_2}$.
Therefore, $x_{N_2} = \frac{4}{1 \times 10^5} = 4 \times 10^{-5}$.
Since the amount of dissolved gas is very small, we can approximate the mole fraction as $x_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}}$.
Given $n_{H_2O} = 10 \ mol$, we have $n_{N_2} = x_{N_2} \times n_{H_2O} = (4 \times 10^{-5}) \times 10 = 4 \times 10^{-4} \ mol$.

(C) According to Henry's law, $P = K_H \times x$, where $x$ is the mole fraction of $CO_2$.
Given $P = 2 \,atm = 2 \times 1.01325 \times 10^5 \,Pa = 2.0265 \times 10^5 \,Pa$.
$K_H = 1.67 \times 10^8 \,Pa$.
$x = \frac{P}{K_H} = \frac{2.0265 \times 10^5}{1.67 \times 10^8} \approx 1.213 \times 10^{-3}$.
Since the amount of $CO_2$ is small, the number of moles of water in $1 \,L$ $(1000 \,g)$ is $n_{H_2O} = \frac{1000}{18} \approx 55.55 \,mol$.
$x = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}}$.
$n_{CO_2} = x \times n_{H_2O} = 1.213 \times 10^{-3} \times 55.55 \approx 0.0674 \,mol$.
Mass of $CO_2 = n_{CO_2} \times \text{Molar mass} = 0.0674 \times 44 \approx 2.96 \,g$.
Rounding to the nearest provided option, the correct answer is $2.9 \,g$.

(B) According to Henry's law, $p = K_H \times \chi_{CO_2}$.
Given: $p = 5 \ bar$, $K_H = 1.67 \ kbar = 1670 \ bar$.
Since the amount of $CO_2$ is small, the mole fraction $\chi_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}}$.
For $1000 \ mL$ of water, the number of moles of water $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} \approx 55.5 \ mol$.
Substituting the values: $5 = 1670 \times \frac{n_{CO_2}}{55.5}$.
$n_{CO_2} = \frac{5 \times 55.5}{1670} \approx 0.166 \ mol \approx 0.167 \ mol$.

(B, C) The solubility of $I_2$ is governed by the principle of 'like dissolves like'.
$I_2$ is a non-polar covalent molecule, making it poorly soluble in polar solvents like water.
However, $I_2$ dissolves in an aqueous solution of $KI$ due to the formation of the soluble triiodide complex ion: $I_2 + I^- \rightarrow I_3^-$.
Additionally, $I_2$ is freely soluble in non-polar organic solvents like $CCl_4$ because both are non-polar in nature.
Therefore, both statements $B$ and $C$ are correct.

(B) The solubility of silver halides in water depends on the lattice energy and hydration energy.
$AgF$ is highly soluble in water due to its high hydration energy.
For the other silver halides ($AgCl$, $AgBr$, $AgI$), the lattice energy decreases as the size of the halide ion increases $(Cl^- < Br^- < I^-)$.
However, the decrease in hydration energy is more significant than the decrease in lattice energy as we move from $Cl^-$ to $I^-$.
Therefore, the solubility decreases in the order: $AgCl > AgBr > AgI$.
Combining these, the overall order of solubility is $AgI < AgBr < AgCl < AgF$.

(D) According to Henry's Law, $P_{N_2} = K_H \cdot X_{N_2}$.
First, calculate the partial pressure of $N_2$ in air: $P_{N_2} = \text{mole fraction of } N_2 \times \text{Total pressure} = 0.8 \times 10 \ atm = 8 \ atm$.
Convert the partial pressure to $mm \ Hg$: $P_{N_2} = 8 \ atm \times 760 \ mm \ Hg/atm = 6080 \ mm \ Hg$.
Now, use Henry's Law to find the mole fraction $X_{N_2}$ in water: $X_{N_2} = \frac{P_{N_2}}{K_H} = \frac{6080}{6.5 \times 10^7}$.
$X_{N_2} = 9.35 \times 10^{-5}$.

(B) Statement $I$ is true because Henry's law states that $p = K_{H}x$, where $K_{H}$ is a constant for a given gas-solvent system at a constant temperature, especially in the range where the solution behaves as an ideally dilute solution.
Statement $II$ is false because $K_{H}$ depends on the nature of the gas and the nature of the solvent. Therefore, $K_{H}$ will differ for the same solute in different solvents.

(C) According to Henry's law, the partial pressure of a gas $(P_g)$ is directly proportional to its mole fraction $(X_g)$ in the solution: $P_g = K_H \cdot X_g$.
Taking the logarithm on both sides: $\log(P_g) = \log(K_H \cdot X_g) = \log(K_H) + \log(X_g)$.
This equation is of the form $y = mx + c$, where $y = \log(P_g)$, $x = \log(X_g)$, the slope $m = 1$, and the intercept $c = \log(K_H)$.
Since the slope is $1$ (positive) and there is a positive intercept $\log(K_H)$, the plot is a straight line with a positive slope that does not pass through the origin. This corresponds to Plot $C$.

(A) $n$-Octane is a non-polar solvent. According to the principle 'like dissolves like',non-polar solutes are more soluble in non-polar solvents.
The polarity order of the given compounds is: $KCl$ (ionic) > $CH_3OH$ (polar) > $CH_3CN$ (polar) > Cyclohexane (non-polar).
Since solubility in a non-polar solvent ($n$-octane) increases as the polarity of the solute decreases, the order of solubility is:
$KCl < CH_3OH < CH_3CN < \text{Cyclohexane}$.
Thus, the correct order is $II < III < IV < I$.

(D) $1$. Solubility of a gas in water depends on the nature of the gas and its interaction with water.
$2$. $NH_3$ and $HCl$ are polar gases that react with water or form strong hydrogen bonds, making them highly soluble.
$3$. $CO_2$ is slightly polar and reacts to some extent with water to form carbonic acid $(H_2CO_3)$, making it moderately soluble.
$4$. $N_2$ is a non-polar, diatomic molecule with very weak van der Waals forces, resulting in very low solubility in water.
$5$. Therefore, $N_2$ is the least soluble gas among the given options.

(D) $1$. Solubility of a gas in water depends on the nature of the gas and its interaction with water.
$2$. $HCl$, $NH_3$, and $CO_2$ are polar or capable of reacting with water (e.g., $NH_3$ forms $NH_4OH$, $CO_2$ forms $H_2CO_3$, $HCl$ ionizes).
$3$. $O_2$ is a non-polar diatomic molecule with very weak van der Waals forces, making it the least soluble in water compared to the others.
$4$. Therefore, $O_2$ is the least soluble gas.

(A) According to Henry's Law, the solubility $(S)$ of a gas is directly proportional to its partial pressure $(P)$: $S = k_H \times P$.
Step $1$: Calculate Henry's constant $(k_H)$ using the initial conditions:
$k_H = \frac{S_1}{P_1} = \frac{6.85 \times 10^{-4} \text{ mol L}^{-1}}{1 \text{ bar}} = 6.85 \times 10^{-4} \text{ mol L}^{-1} \text{ bar}^{-1}$.
Step $2$: Calculate the new solubility $(S_2)$ at $P_2 = 0.70 \text{ bar}$:
$S_2 = k_H \times P_2 = (6.85 \times 10^{-4} \text{ mol L}^{-1} \text{ bar}^{-1}) \times (0.70 \text{ bar}) = 4.795 \times 10^{-4} \text{ mol L}^{-1} \approx 4.80 \times 10^{-4} \text{ mol L}^{-1}$.

(A) According to Henry's Law, the concentration of a dissolved gas is given by the formula: $C = K_H \times P$
Given:
Partial pressure, $P = 0.18 \text{ atm}$
Henry's Law constant, $K_H = 0.15 \text{ mol dm}^{-3} \text{atm}^{-1}$
Calculation:
$C = 0.15 \text{ mol dm}^{-3} \text{atm}^{-1} \times 0.18 \text{ atm}$
$C = 0.027 \text{ mol dm}^{-3}$
Since $1 \text{ mol dm}^{-3} = 1 \text{ M}$, the concentration is $0.027 \text{ M}$.

(A) Henry's law is applicable to gases that do not undergo any chemical reaction with the solvent. $NH_3$ and $CO_2$ react with water to form $NH_4OH$ and $H_2CO_3$ respectively. Therefore, these gases do not obey Henry's law.

(C) According to Henry's law, the solubility $(S)$ of a gas is given by the formula: $S = K_H \times P$
Given:
$K_H = 3.0 \times 10^{-2} \text{ mol dm}^{-3} \text{ atm}^{-1}$
$P = 3 \text{ atm}$
Substituting the values:
$S = (3.0 \times 10^{-2} \text{ mol dm}^{-3} \text{ atm}^{-1}) \times (3 \text{ atm})$
$S = 9.0 \times 10^{-2} \text{ mol dm}^{-3}$
$S = 0.09 \text{ M}$
Therefore, the solubility is $0.09 \text{ M}$.

(A) $1$. Soft drinks are bottled under high pressure to increase the solubility of $CO_2$ gas in the liquid (Henry's Law).
$2$. When the bottle is opened, the internal pressure drops to atmospheric pressure.
$3$. Due to the decrease in pressure, the solubility of the dissolved $CO_2$ gas decreases (Statement $A$ is correct).
$4$. The bottle releases the excess internal pressure (Statement $B$ is correct).
$5$. The dissolved gas escapes rapidly, causing effervescence (Statement $C$ is correct).
$6$. External pressure remains constant; it does not increase (Statement $D$ is incorrect).
$7$. Therefore, $(A)$, $(B)$, and $(C)$ are correct.

(B) $1$. $A$ solution is a homogeneous mixture of two or more substances.
$2$. In a solution, the component present in a smaller amount is the solute, and the component present in a larger amount is the solvent.
$3$. Sea water is a solution where salts (solids) are dissolved in water (liquid).
$4$. Hydrogen in Palladium is a gas in solid solution.
$5$. Gels are colloids where a liquid is dispersed in a solid.
$6$. Amalgams are solutions of metals (solids) in mercury (liquid), but in common classification, sea water is the standard example of a solid solute in a liquid solvent.

(D) Step $1$: Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Step $2$: However, Henry's law is only applicable to gases that do not undergo chemical reactions with the solvent.
Step $3$: Gases like $NH_3$ and $CO_2$ react with water (e.g., $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$), therefore they do not strictly obey Henry's law.
Step $4$: Thus, statement $(D)$ is not true.

(A) Brass is an alloy of copper and zinc. Since both copper and zinc are solids at room temperature, brass is a homogeneous mixture of two solids. Therefore, it is classified as a solid-in-solid solution.