Characteristics and Measurable properties of gases Questions in English

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
For two gases $A$ and $B$,the ratio of their rates of diffusion is given by: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Given $M_A = 36$ and $M_B = 64$.
Substituting the values: $\frac{r_A}{r_B} = \sqrt{\frac{64}{36}} = \frac{8}{6} = \frac{4}{3}$.
Therefore,the ratio $r_A : r_B$ is $4:3$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
From the ideal gas equation,$PV = nRT = \frac{w}{M}RT$.
Rearranging gives $PM = \frac{w}{V}RT = dRT$,where $d$ is the density.
Thus,$M = \frac{dRT}{P}$.
At a constant temperature,$M \propto \frac{d}{P}$.
Substituting this into the rate equation: $r \propto \frac{1}{\sqrt{d/P}} = \sqrt{\frac{P}{d}}$.

(B) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
For a given gas,$M$ and $R$ are constant,so $d \propto \frac{P}{T}$.
To maximize density,we need the highest ratio of $\frac{P}{T}$.
Calculating $\frac{P}{T}$ for each option:
$A) \frac{1 \, atm}{273 \, K} \approx 0.00366 \, atm/K$
$B) \frac{2 \, atm}{273 \, K} \approx 0.00732 \, atm/K$
$C) \frac{1 \, atm}{546 \, K} \approx 0.00183 \, atm/K$
$D) \frac{3 \, atm}{546 \, K} \approx 0.00549 \, atm/K$
Comparing the values,option $B$ gives the maximum ratio.

(B) Dalton's law of partial pressure is applicable only to non-reacting gas mixtures.
In the mixture $NH_3 + HCl + HBr$,$NH_3$ reacts with $HCl$ and $HBr$ to form solid ammonium salts ($NH_4Cl$ and $NH_4Br$).
Therefore,this mixture does not follow the law.

(B) According to Dalton's Law of Partial Pressures,the total pressure is the sum of partial pressures of individual gases.
Since the number of molecules of $N_2$ and $O_2$ are equal,their mole fractions are equal.
Therefore,the partial pressure of each gas is half of the total pressure.
$P_{total} = P_{N_2} + P_{O_2} = 650 \ mm$
Since $P_{N_2} = P_{O_2}$,we have $2 \times P_{O_2} = 650 \ mm$.
Thus,$P_{O_2} = \frac{650}{2} \ mm = 325 \ mm$.
If $N_2$ is removed,only $O_2$ remains,and the pressure in the vessel will be the partial pressure of $O_2$,which is $325 \ mm$ or $\frac{650}{2} \ mm$.

(A) The molar mass of both $CO$ and $N_2$ is $28 \, g/mol$.
Since the masses are equal,the number of moles $(n = \frac{mass}{molar \, mass})$ for both gases will be equal.
According to Dalton's Law of Partial Pressures,the partial pressure of a gas is directly proportional to its mole fraction $(P_i = x_i \times P_{total})$.
Since the number of moles of both gases is equal,their mole fractions are equal $(x_{CO} = x_{N_2} = 0.5)$.
Therefore,the partial pressures are equal: $P_{CO} = P_{N_2}$.

(D) Total pressure $P_{total} = P_{N_2} + P_{O_2} + P_{H_2O} = 640 \, torr$.
Given $P_{H_2O} = 40 \, torr$,so the pressure of dry air is $P_{dry} = P_{N_2} + P_{O_2} = 640 - 40 = 600 \, torr$.
Since partial pressure is proportional to the mole fraction,the ratio of partial pressures $P_{N_2} : P_{O_2}$ is equal to the molar ratio $3:1$.
Let $P_{N_2} = 3x$ and $P_{O_2} = x$.
$3x + x = 600 \implies 4x = 600 \implies x = 150 \, torr$.
Therefore,$P_{N_2} = 3 \times 150 = 450 \, torr$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_A}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_A}}$.
Given that $r_A = 6 \times r_{H_2}$ and the molar mass of hydrogen $M_{H_2} = 2 \ g/mol$.
Substituting the values: $6 = \sqrt{\frac{2}{M_A}}$.
Squaring both sides: $36 = \frac{2}{M_A}$.
Therefore,$M_A = \frac{2}{36} = \frac{1}{18} \ g/mol$.
Wait,re-evaluating the standard interpretation: If the rate of $A$ is $6$ times that of $H_2$,then $M_A = M_{H_2} / 6^2 = 2 / 36 = 1/18$. However,if the question implies $H_2$ is $6$ times faster than $A$,then $M_A = 2 \times 6^2 = 72$. Given the options,the intended calculation is $M_A = 2 \times 6^2 = 72$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the rates of diffusion of $SO_2$ and $O_2$ is given by: $\frac{r_{SO_2}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{SO_2}}}$.
The molar mass of $SO_2$ is $64 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$.
Substituting these values: $\frac{r_{SO_2}}{r_{O_2}} = \sqrt{\frac{32}{64}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio is $1 : \sqrt{2}$.

(A) According to Graham's law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molecular mass $M$: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Since $r = \frac{V}{t}$ and the volumes $V$ are equal,$\frac{r_A}{r_B} = \frac{t_B}{t_A}$.
Substituting the given values: $\frac{200}{150} = \sqrt{\frac{36}{M_A}}$.
Simplifying the ratio: $\frac{4}{3} = \sqrt{\frac{36}{M_A}}$.
Squaring both sides: $\frac{16}{9} = \frac{36}{M_A}$.
Solving for $M_A$: $M_A = \frac{36 \times 9}{16} = \frac{324}{16} = 20.25$.

(B) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M_W$: $\frac{r_1}{r_2} = \sqrt{\frac{M_{W_2}}{M_{W_1}}}$.
Since rate $r = \frac{V}{t}$,for the same volume $V$,the expression becomes $\frac{t_2}{t_1} = \sqrt{\frac{M_{W_1}}{M_{W_2}}}$.
Given that the time taken by the gas $(t_1)$ is $3$ times the time taken by helium $(t_2)$,we have $\frac{t_1}{t_2} = 3$.
Substituting the values: $3 = \sqrt{\frac{M_{W_1}}{4}}$.
Squaring both sides: $9 = \frac{M_{W_1}}{4}$.
Therefore,$M_{W_1} = 9 \times 4 = 36 \ u$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ for gases at the same volume and pressure: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Since $r = \frac{V}{t}$ and $V_A = V_B$,the equation becomes $\frac{t_B}{t_A} = \sqrt{\frac{M_B}{M_A}}$.
Given $t_A = 20 \ s$,$t_B = 10 \ s$,and $M_A = 49 \ u$,we substitute these values:
$\frac{10}{20} = \sqrt{\frac{M_B}{49}}$.
$\frac{1}{2} = \frac{\sqrt{M_B}}{7}$.
$\sqrt{M_B} = \frac{7}{2} = 3.5$.
$M_B = (3.5)^2 = 12.25 \ u$.

(A) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is proportional to its mole fraction in the mixture.
Given that the moles of $CO$ and $N_2$ are equal,let $n_{CO} = n_{N_2} = n$.
The mole fraction of $N_2$ is given by $x_{N_2} = \frac{n_{N_2}}{n_{CO} + n_{N_2}} = \frac{n}{n + n} = \frac{n}{2n} = 0.5$.
The partial pressure of $N_2$ is calculated as $p_{N_2} = x_{N_2} \times P_{total}$.
Given $P_{total} = 1 \ atm$,we have $p_{N_2} = 0.5 \times 1 \ atm = 0.5 \ atm$.

(C) Intermolecular forces are the forces of attraction between molecules.
When these forces increase,it becomes harder for molecules to escape into the vapour phase.
Consequently,the tendency of a liquid to evaporate decreases,which leads to a decrease in vapour pressure.
Conversely,properties like boiling point,enthalpy of vapourisation,and viscosity increase as intermolecular forces increase because more energy is required to overcome these stronger attractions.

(B) Let the mass of each gas be $32 \ g$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{32 \ g}{2 \ g/mol} = 16 \ mol$.
Number of moles of $O_2$ $(n_{O_2})$ = $\frac{32 \ g}{32 \ g/mol} = 1 \ mol$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{32 \ g}{16 \ g/mol} = 2 \ mol$.
Total moles = $16 + 1 + 2 = 19 \ mol$.
Mole fraction of $CH_4$ $(x_{CH_4})$ = $\frac{n_{CH_4}}{\text{Total moles}} = \frac{2}{19}$.
Partial pressure of $CH_4$ $(P_{CH_4})$ = $x_{CH_4} \times P_{total} = \frac{2}{19} \times 380 \ torr = 40 \ torr$.

(A) According to Graham's Law of diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its density $(d)$ or molar mass $(M)$.
Additionally,the rate of diffusion is directly proportional to the pressure $(P)$ applied.
Therefore,the relationship is given by $r \propto \frac{P}{\sqrt{M}}$.
Since density $(d)$ is directly proportional to molar mass $(M)$ at a constant temperature,$r \propto \frac{P}{\sqrt{d}}$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r \propto \frac{W}{t \sqrt{M}}$.
Therefore,$\frac{r_{O_{2}}}{r_{N_{2}}} = \frac{W_{O_{2}} / t_{O_{2}}}{W_{N_{2}} / t_{N_{2}}} = \sqrt{\frac{M_{N_{2}}}{M_{O_{2}}}}$
Given: $W_{O_{2}} = 3.2 \, g$,$t_{O_{2}} = 10 \, minute$,$M_{O_{2}} = 32 \, g/mol$,$W_{N_{2}} = 2.8 \, g$,$M_{N_{2}} = 28 \, g/mol$.
Substituting the values: $\frac{3.2 / 10}{2.8 / t_{N_{2}}} = \sqrt{\frac{28}{32}}$
$\frac{3.2 \times t_{N_{2}}}{28} = \sqrt{0.875} \approx 0.9354$
$t_{N_{2}} = \frac{0.9354 \times 28}{3.2} \approx 8.185 \, minute \approx 8.2 \, minute$.

(B) Viscosity is a measure of the resistance of a liquid to flow,which arises due to the attractive forces between molecules.
As the temperature increases,the average kinetic energy of the liquid molecules increases.
This increased kinetic energy allows the molecules to overcome the intermolecular attractive forces that hold them together.
Consequently,the liquid flows more easily,leading to a decrease in viscosity.

(A) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is given by the product of its mole fraction and the total pressure of the mixture.
Since the mixture contains equal moles of $CO$ and $N_2$,the mole fraction of $N_2$ $(X_{N_2})$ is $\frac{1}{1+1} = 0.5$.
Given the total pressure $(P_T) = 1 \ atm$.
Therefore,the partial pressure of $N_2$ is $P_{N_2} = X_{N_2} \times P_T = 0.5 \times 1 = 0.5 \ atm$.

(A) According to Graham's Law of diffusion,the rate of diffusion $(r)$ of a gas is directly proportional to its pressure $(p)$ and inversely proportional to the square root of its density $(d)$.
$r \propto \frac{p}{\sqrt{d}}$
This implies that $r$ is inversely proportional to $\frac{\sqrt{d}}{p}$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Given $M_{O_2} = 32 \, g/mol$ and $M_{SO_2} = 64 \, g/mol$.
The rate $r = \frac{V}{t}$,so $\frac{V_{O_2}/t_{O_2}}{V_{SO_2}/t_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{O_2}}}$.
Substituting the values: $\frac{80/2}{V_{SO_2}/3} = \sqrt{\frac{64}{32}}$.
$40 \times \frac{3}{V_{SO_2}} = \sqrt{2}$.
$V_{SO_2} = \frac{120}{\sqrt{2}} \, mL$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ for the same volume of gas: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Since $r = \frac{V}{t}$ and the volume $V$ is the same,$\frac{r_A}{r_B} = \frac{t_B}{t_A}$.
Given $t_A = 20 \ s$,$t_B = 10 \ s$,and $M_A = 49 \ u$,we have:
$\frac{10}{20} = \sqrt{\frac{M_B}{49}}$
$\frac{1}{2} = \sqrt{\frac{M_B}{49}}$
Squaring both sides:
$\frac{1}{4} = \frac{M_B}{49}$
$M_B = \frac{49}{4} = 12.25 \ u$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $Mw$: $r \propto \frac{1}{\sqrt{Mw}}$.
Therefore,$\frac{r_x}{r_y} = \sqrt{\frac{Mw_y}{Mw_x}}$.
Given $r_x = 10 \, mL/s$ and $r_y = 40 \, mL/s$,we have:
$\frac{10}{40} = \sqrt{\frac{Mw_y}{Mw_x}}$
$\frac{1}{4} = \sqrt{\frac{Mw_y}{Mw_x}}$
Squaring both sides:
$\frac{1}{16} = \frac{Mw_y}{Mw_x}$
Thus,the ratio of their molar masses $\frac{Mw_x}{Mw_y} = \frac{16}{1}$ or $16 : 1$.

(B) From the ideal gas equation,$PV = nRT$,we have $PV = (\frac{m}{M_w})RT$.
Rearranging gives $P = (\frac{m}{V}) \frac{RT}{M_w}$,so $P = d \frac{RT}{M_w}$,where $d$ is the density.
Thus,$d = \frac{PM_w}{RT}$.
At constant temperature,$d \propto P$.
At high altitudes,the pressure of air is low,which means the density of air is also low.
Therefore,the statement that the density of air is high at high altitude is incorrect.

(D) According to Graham's Law,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
First,calculate the molar masses of the given gases:
$M(CO_2) = 12 + 2 \times 16 = 44 \ g/mol$
$M(SO_2) = 32 + 2 \times 16 = 64 \ g/mol$
$M(SO_3) = 32 + 3 \times 16 = 80 \ g/mol$
$M(PCl_3) = 31 + 3 \times 35.5 = 137.5 \ g/mol$
Since the rate of diffusion is inversely proportional to the square root of the molar mass,the gas with the lowest molar mass will have the highest rate of diffusion.
The order of molar masses is: $CO_2 (44) < SO_2 (64) < SO_3 (80) < PCl_3 (137.5)$.
Therefore,the order of the rate of diffusion is: $CO_2 > SO_2 > SO_3 > PCl_3$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.
For two gases,the ratio is $\frac{r_1}{r_2} = \frac{V_1}{t_1} \times \frac{t_2}{V_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $V_1 = 80 \ mL$,$t_1 = 2 \ min$,$M_1 (O_2) = 32 \ g/mol$,$t_2 = 3 \ min$,$M_2 (SO_2) = 64 \ g/mol$.
Substituting the values: $\frac{80}{2} \times \frac{3}{V_2} = \sqrt{\frac{64}{32}}$.
$40 \times \frac{3}{V_2} = \sqrt{2}$.
$V_2 = \frac{120}{\sqrt{2}} \ mL$.

(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M_w$:
$\frac{r_A}{r_B} = \sqrt{\frac{M_{wB}}{M_{wA}}}$
Since rate $r = \frac{V}{t}$,and volumes are equal $(V_A = V_B)$:
$\frac{t_B}{t_A} = \sqrt{\frac{M_{wB}}{M_{wA}}}$
Substituting the given values ($t_A = 20 \ s$,$t_B = 10 \ s$,$M_{wA} = 80$):
$\frac{10}{20} = \sqrt{\frac{M_{wB}}{80}}$
$0.5 = \sqrt{\frac{M_{wB}}{80}}$
Squaring both sides:
$0.25 = \frac{M_{wB}}{80}$
$M_{wB} = 0.25 \times 80 = 20$

(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
The rate of diffusion is defined as $r = \frac{V}{t}$,where $V$ is the volume and $t$ is the time.
Thus,$\frac{r_{SO_2}}{r_{O_2}} = \frac{V_{SO_2} / t_{SO_2}}{V_{O_2} / t_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{SO_2}}}$.
Given: $V_{SO_2} = 20 \ dm^3$,$t_{SO_2} = 60 \ s$,$t_{O_2} = 30 \ s$.
Molar mass of $SO_2 = 32 + (2 \times 16) = 64 \ g/mol$.
Molar mass of $O_2 = 2 \times 16 = 32 \ g/mol$.
Substituting the values: $\frac{20 / 60}{V_{O_2} / 30} = \sqrt{\frac{32}{64}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
$\frac{1/3}{V_{O_2}/30} = \frac{1}{\sqrt{2}} \implies \frac{10}{V_{O_2}} = \frac{1}{1.414}$.
$V_{O_2} = 10 \times 1.414 = 14.14 \ dm^3 \approx 14.1 \ dm^3$.

(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is given by $r = \frac{w}{M \times t}$,where $w$ is the mass,$M$ is the molar mass,and $t$ is the time.
For the same volume of gas,the ratio of rates is $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.
Substituting the values: $\frac{w_1 / (M_1 \times t_1)}{w_2 / (M_2 \times t_2)} = \sqrt{\frac{M_2}{M_1}}$.
Given: $w_{O_2} = 3.2 \ g$,$M_{O_2} = 32 \ g/mol$,$t_{O_2} = 10 \ min$,$w_{N_2} = 2.8 \ g$,$M_{N_2} = 28 \ g/mol$.
$\frac{3.2 / (32 \times 10)}{2.8 / (28 \times t_{N_2})} = \sqrt{\frac{28}{32}}$.
$\frac{0.1 / 10}{0.1 / t_{N_2}} = \sqrt{0.875}$.
$\frac{t_{N_2}}{10} = 0.9354$.
$t_{N_2} = 9.35 \ min \approx 9.3 \ min$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $Mw$: $r \propto \frac{1}{\sqrt{Mw}}$.
Therefore,the ratio of rates is given by: $\frac{r_{x}}{r_{y}} = \sqrt{\frac{(Mw)_{y}}{(Mw)_{x}}}$.
Given $r_{x} = 10 \ mL/sec$ and $r_{y} = 40 \ mL/sec$,we have: $\frac{10}{40} = \sqrt{\frac{(Mw)_{y}}{(Mw)_{x}}}$.
Squaring both sides: $(\frac{1}{4})^2 = \frac{(Mw)_{y}}{(Mw)_{x}}$,which gives $\frac{1}{16} = \frac{(Mw)_{y}}{(Mw)_{x}}$.
Thus,the ratio of molar masses $(Mw)_{x} : (Mw)_{y}$ is $16:1$.

(D) Liquid ammonia exhibits hydrogen bonding,but it possesses a very high vapour pressure at room temperature.
Because the vapour pressure is significantly high,the ammonia will rapidly evaporate and potentially cause the bottle to burst or spray upon opening.
Cooling the bottle in ice decreases the vapour pressure,allowing the liquid to remain in a stable state.
Therefore,it is recommended to cool the bottle before opening it to ensure safety.

(A) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the rate of diffusion of $O_2$ to $H_2$ is given by: $\frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$.
The molar mass of $O_2$ is $32 \ g/mol$ and the molar mass of $H_2$ is $2 \ g/mol$.
Substituting these values: $\frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,the ratio is $1 : 4$.

(A) Dalton's law of partial pressure is applicable only to a mixture of non-reacting gases.
In the case of $NH_{3(g)}$ and $HCl_{(g)}$,they react to form solid ammonium chloride $(NH_4Cl_{(s)})$ according to the reaction: $NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}$.
Since they react with each other,Dalton's law is not applicable to this mixture.

(B) In the gaseous state,the particles are far apart and move randomly at high speeds due to high kinetic energy.
This high kinetic energy,which is a form of thermal energy,overcomes the weak intermolecular forces of attraction between the gas molecules.
Therefore,for the gaseous state,the thermal energy is much greater than the molecular forces of attraction $(Thermal \ energy >> Molecular \ attraction)$.

(A) According to Dalton's Law,the partial pressure $P_i$ of a gas is proportional to its mole fraction $x_i$ $(P_i = x_i P_{total})$.
Since $P_i = \frac{n_i RT}{V}$,for equal masses $W$ of $CO$ and $N_2$:
$n_{CO} = \frac{W}{M_{CO}} = \frac{W}{28}$ and $n_{N_2} = \frac{W}{M_{N_2}} = \frac{W}{28}$.
Since the molar masses of $CO$ $(12+16=28 \ g/mol)$ and $N_2$ $(14 \times 2=28 \ g/mol)$ are equal,the number of moles $n$ is the same for both gases.
Therefore,$\frac{P_{CO}}{P_{N_2}} = \frac{n_{CO}}{n_{N_2}} = \frac{W/28}{W/28} = 1$.
Thus,$P_{CO} = P_{N_2}$.

(A) The density $(d)$ of a gas is given by the formula: $d = \frac{PM}{RT}$.
At $STP$,the pressure $(P) = 1 \ atm$,the molar mass $(M)$ of $COCl_2 = 12 + 16 + 2 \times 35.5 = 99 \ g \ mol^{-1}$,and the molar volume is $22.4 \ L \ mol^{-1}$.
Thus,$d = \frac{M}{V_m} = \frac{99 \ g \ mol^{-1}}{22.4 \ L \ mol^{-1}} = 4.42 \ g \ L^{-1}$.

(C) According to Graham's Law of Diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
Since $H_2$ has the lowest molar mass $(M = 2 \ g/mol)$ compared to $He$ $(M = 4 \ g/mol)$,$N_2$ $(M = 28 \ g/mol)$,and $O_2$ $(M = 32 \ g/mol)$,it will diffuse the fastest.
Therefore,$H_2$ will fill the tube first.

(B) The number of moles of $H_2$ is $n_{H_2} = \frac{1 \ g}{2 \ g/mol} = 0.5 \ mol$.
The number of moles of $He$ is $n_{He} = \frac{1 \ g}{4 \ g/mol} = 0.25 \ mol$.
The total number of moles is $n_{total} = 0.5 + 0.25 = 0.75 \ mol$.
The mole fraction of $He$ is $x_{He} = \frac{n_{He}}{n_{total}} = \frac{0.25}{0.75} = \frac{1}{3}$.
The partial pressure of $He$ is $P_{He} = x_{He} \times P_{total} = \frac{1}{3} \times 1.2 \times 10^3 \ Pa = 0.4 \times 10^3 \ Pa = 400 \ Pa$.

(D) Let the mass of each gas be $m \ g$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{m}{2}$.
Number of moles of $He$ $(n_{He})$ = $\frac{m}{4}$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{m}{16}$.
Total moles $(n_{total})$ = $\frac{m}{2} + \frac{m}{4} + \frac{m}{16} = \frac{8m + 4m + m}{16} = \frac{13m}{16}$.
Mole fraction of $H_2$ $(x_{H_2})$ = $\frac{n_{H_2}}{n_{total}} = \frac{m/2}{13m/16} = \frac{1}{2} \times \frac{16}{13} = \frac{8}{13}$.
Partial pressure of $H_2$ $(P_{H_2})$ = $x_{H_2} \times P_{total} = \frac{8}{13} \times 1300 \ torr = 800 \ torr$.

(C) Dalton's Law of Partial Pressures is applicable only to non-reacting gases.
$NH_3$ (ammonia) and $HCl$ (hydrogen chloride) react with each other to form solid ammonium chloride $(NH_4Cl)$:
$NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$
Since they react,they do not follow Dalton's law.

(C) The partial pressure of a gas is given by the product of its mole fraction and the total pressure $(P_i = x_i \times P_{total})$.
Given the mole ratio $n_A : n_B = 2 : 5$,the total number of parts is $2 + 5 = 7$.
The mole fraction of gas $B$ is $x_B = \frac{5}{7}$.
The total pressure $P_{total} = 3.5 \times 10^5 \ Pa$.
Therefore,the partial pressure of $B$ is $P_B = \frac{5}{7} \times 3.5 \times 10^5 \ Pa$.
$P_B = 5 \times 0.5 \times 10^5 \ Pa = 2.5 \times 10^5 \ Pa$.

(D) According to Dalton's Law of Partial Pressures,the total pressure $P_{total} = P_{N_2} + P_{O_2} + P_{H_2O}$.
Given $P_{total} = 640 \ torr$ and $P_{H_2O} = 40 \ torr$.
Therefore,the sum of partial pressures of dry gases is $P_{N_2} + P_{O_2} = 640 - 40 = 600 \ torr$.
Since the molar ratio of $N_2$ to $O_2$ is $3 : 1$,the partial pressures are proportional to their mole fractions.
Let $P_{N_2} = 3x$ and $P_{O_2} = x$.
Then $3x + x = 600 \ torr$,which gives $4x = 600 \ torr$,so $x = 150 \ torr$.
The partial pressure of $N_2$ is $P_{N_2} = 3x = 3 \times 150 = 450 \ torr$.

(A) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r_1 / r_2 = \sqrt{M_2 / M_1}$.
Since the time $t$ is the same for both gases,the rate is proportional to the amount of gas effused: $n_1 / n_2 = \sqrt{M_2 / M_1}$.
Given: $n(O_2) = 0.48 \ g / 32 \ g/mol = 0.015 \ mol$.
Molar mass of $O_2 = 32 \ g/mol$,Molar mass of $CO_2 = 44 \ g/mol$.
$0.015 / n(CO_2) = \sqrt{44 / 32} = \sqrt{1.375} \approx 1.1726$.
$n(CO_2) = 0.015 / 1.1726 \approx 0.01279 \ mol$.
At $STP$,$1 \ mol$ of gas occupies $22400 \ mL$.
Volume of $CO_2 = 0.01279 \ mol \times 22400 \ mL/mol \approx 286.5 \ mL$.

(D) Let the mass of each gas be $m \, g$.
Number of moles of the lighter gas $(M_1 = 4)$ is $n_1 = \frac{m}{4}$.
Number of moles of the heavier gas $(M_2 = 40)$ is $n_2 = \frac{m}{40}$.
Total moles $n_{total} = n_1 + n_2 = \frac{m}{4} + \frac{m}{40} = \frac{10m + m}{40} = \frac{11m}{40}$.
Mole fraction of the lighter gas $(x_1)$ = $\frac{n_1}{n_{total}} = \frac{m/4}{11m/40} = \frac{m}{4} \times \frac{40}{11m} = \frac{10}{11}$.
Partial pressure of the lighter gas = $x_1 \times P_{total} = \frac{10}{11} \times 1.1 \, atm = 1.0 \, atm$.