Characteristics and Measurable properties of gases Questions in English

(A) The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the external atmospheric pressure.
At higher altitudes,the atmospheric pressure is lower than the standard pressure of $76 \ cm \ Hg$.
Therefore,the liquid reaches its boiling point at a lower temperature.

(B) Applying the Clausius-Clapeyron equation:
$\log \frac{P_2}{P_1} = \frac{\Delta H_{vap}}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given: $P_2 = 760 \ mm$,$T_2 = 373 \ K$,$P_1 = 23 \ mm$,$\Delta H_{vap} = 40656 \ J/mol$,$R = 8.314 \ J/K \cdot mol$.
Substituting the values:
$\log \frac{760}{23} = \frac{40656}{2.303 \times 8.314} \left[ \frac{373 - T_1}{373 T_1} \right]$
$1.519 = 2123.5 \times \left[ \frac{373 - T_1}{373 T_1} \right]$
$0.000715 = \frac{373 - T_1}{373 T_1}$
$0.2667 T_1 = 373 - T_1$
$1.2667 T_1 = 373$
$T_1 \approx 294.4 \ K$.

(D) Viscosity is a measure of a fluid's resistance to flow. It is primarily determined by the intermolecular forces within the liquid and the temperature of the liquid.
Among the given options,the viscosity of a liquid is related to its density,as both properties are influenced by the strength of intermolecular forces and the molecular structure of the substance.
Therefore,the correct option is $D$.

(D) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r_1 / r_2 = \sqrt{M_2 / M_1}$.
Given: Molar mass of gas $(M_1 = 128)$,Molar mass of oxygen $(M_2 = 32)$.
Substituting the values: $r_{gas} / r_{O_2} = \sqrt{32 / 128} = \sqrt{1 / 4} = 0.5$.
Therefore,the rate of diffusion of the gas is $0.5$ times that of oxygen.

(B) As altitude increases,the atmospheric pressure decreases,which leads to a decrease in the density of air. Consequently,the availability of oxygen for breathing becomes insufficient at high altitudes,necessitating the use of oxygen cylinders.

(D) Dalton's law of partial pressure is applicable only to non-reacting gas mixtures.
$NH_3$ and $HCl$ react with each other to form solid $NH_4Cl$ $(NH_3(g) + HCl(g) \rightarrow NH_4Cl(s))$.
Therefore,they do not obey Dalton's law of partial pressure.

(A) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of its molecular weight $(M)$: $r_1 / r_2 = \sqrt{M_2 / M_1}$.
Given: Molecular weight of gas $(M_1)$ = $98$,Molecular weight of hydrogen $(M_2)$ = $2$.
Substituting the values: $r_{gas} / r_{H_2} = \sqrt{2 / 98} = \sqrt{1 / 49} = 1/7$.
Therefore,the rate of diffusion of the gas is $1/7$ times that of hydrogen.

(A) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r_1 / r_2 = \sqrt{M_2 / M_1}$.
Given: $r_{CH_4} = 2 \times r_X$,$M_{CH_4} = 16 \ g/mol$.
Substituting the values: $2 \times r_X / r_X = \sqrt{M_X / 16}$.
$2 = \sqrt{M_X / 16}$.
Squaring both sides: $4 = M_X / 16$.
$M_X = 4 \times 16 = 64 \ g/mol$.

(D) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reactive gases at a constant temperature and volume is equal to the sum of the partial pressures of the individual gases present in the mixture. Mathematically,$P_{total} = p_1 + p_2 + p_3 + \dots$ where $p_1, p_2, p_3$ are the partial pressures of the individual gases.

(B) According to Dalton's Law of Partial Pressures,the partial pressure $(p_i)$ of a gaseous component in a mixture is given by the product of its mole fraction $(x_i)$ and the total pressure $(P_{total})$.
$p_i = x_i \times P_{total}$
Therefore,the ratio of the partial pressure to the total pressure is:
$\frac{p_i}{P_{total}} = x_i$
Thus,the ratio is equal to the mole fraction of the component.

(D) The root mean square velocity $(u_{rms})$ of an ideal gas is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.
From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,we can write $P = \frac{m}{MV}RT = \frac{dRT}{M}$,where $d$ is the density $(d = m/V)$.
Thus,$\frac{RT}{M} = \frac{P}{d}$.
Substituting this into the $u_{rms}$ formula,we get $u_{rms} = \sqrt{\frac{3P}{d}}$.
Since pressure $(P)$ is constant,$u_{rms} \propto \frac{1}{\sqrt{d}}$.

(D) According to Dalton's Law of Partial Pressures,the total pressure of a gas collected over water is the sum of the pressure of the dry gas and the vapor pressure of water.
$P_{total} = P_{dry} + P_{water \ vapor}$
Given:
$P_{total} = 735 \ torr$
$P_{water \ vapor} = 30 \ torr$
Substituting the values:
$735 = P_{dry} + 30$
$P_{dry} = 735 - 30 = 705 \ torr$
Therefore,the pressure exerted by the dry methane gas is $705 \ torr$.

(C) Graham's law of diffusion states that the rate of diffusion $r$ of a gas is directly proportional to its partial pressure $P$ and inversely proportional to the square root of its molar mass $M$.
Mathematically,$r \propto P / \sqrt{M}$.
Therefore,for two gases $A$ and $B$,the ratio of their rates of diffusion is given by:
$r_A/r_B = (P_A/P_B) \times \sqrt{M_B/M_A} = (P_A/P_B)(M_B/M_A)^{1/2}$.

(B) The spreading of fragrance in a room is due to the movement of gas molecules from a region of higher concentration to a region of lower concentration.
This process is known as $Diffusion$.
$Effusion$ refers to the escape of gas molecules through a tiny hole,which is not the case here.

(A) According to Graham's Law of Effusion,the rate of effusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$ or density $(d)$: $r \propto \frac{1}{\sqrt{d}}$.
Furthermore,the rate of effusion is directly proportional to the pressure $(P)$ applied: $r \propto P$.
Combining these two relationships,we get $r \propto \frac{P}{\sqrt{d}}$.

(C) Step $1$: Calculate the number of moles of each gas.
$n(H_2) = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \ g}{2 \ g/mol} = 1 \ mol$.
$n(SO_2) = \frac{32 \ g}{64 \ g/mol} = 0.5 \ mol$.
Step $2$: Calculate the mole fraction of $H_2$.
$x(H_2) = \frac{n(H_2)}{n(H_2) + n(SO_2)} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{1}{3/2} = 2/3$.
Step $3$: Use Dalton's Law of partial pressure.
$P(H_2) = x(H_2) \times P_{\text{total}} = \frac{2}{3} P_{\text{total}}$.

(A) Let the mass of both $CO$ and $N_2$ be $w \ g$.
The molar mass of $CO$ is $28 \ g/mol$ and the molar mass of $N_2$ is $28 \ g/mol$.
The number of moles is given by $n = \frac{\text{mass}}{\text{molar mass}}$.
$n_{CO} = \frac{w}{28}$ and $n_{N_2} = \frac{w}{28}$.
According to Dalton's Law of partial pressure,$P_i = x_i \times P_{total}$,where $x_i$ is the mole fraction.
Since $n_{CO} = n_{N_2}$,the mole fractions are equal $(x_{CO} = x_{N_2} = 0.5)$.
Therefore,the partial pressures are equal: $P_{CO} = P_{N_2}$.

(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 molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
Comparing the molar masses: $M(H_2) = 2 \ g/mol$,$M(He) = 4 \ g/mol$,$M(N_2) = 28 \ g/mol$,$M(O_2) = 32 \ g/mol$.
Since $H_2$ has the lowest molar mass,it has the highest rate of diffusion.
Therefore,the tube filled with $H_2$ will deflate first.

(A) According to the definition of a mole,$1 \ mol$ of any gas always contains $6.022 \times 10^{23}$ molecules (Avogadro's number).
Since the number of moles is constant $(1 \ mol)$,the number of molecules remains constant regardless of changes in pressure or volume,provided the temperature remains constant.

(B) According to Dalton's Law of Partial Pressures,the partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure.
Since equal amounts (moles) of $CH_4$ and $O_2$ are mixed,let the number of moles of each be $n$.
The total number of moles is $n + n = 2n$.
The mole fraction of $O_2$ is $\frac{n}{2n} = 1/2$.
Therefore,the fraction of total pressure exerted by oxygen is $1/2$.

(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: $r \propto \frac{1}{\sqrt{M}}$.
Since $r = \frac{V}{t}$,for the same volume $V$,$t \propto \sqrt{M}$.
Therefore,$\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$.
Given $t_1 = 5 \, s$ for $H_2$ $(M_1 = 2 \, g/mol)$.
For option $B$: $M_2 = 32 \, g/mol$ $(O_2)$.
$t_2 = t_1 \times \sqrt{\frac{M_2}{M_1}} = 5 \times \sqrt{\frac{32}{2}} = 5 \times \sqrt{16} = 5 \times 4 = 20 \, s$.

(B) Surface tension is the result of cohesive forces between liquid molecules. $As$ the temperature increases,the kinetic energy of the molecules increases,which weakens the intermolecular cohesive forces. Consequently,the surface tension of water decreases with an increase in temperature.

(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For hydrogen $(H_2)$,$M_1 = 2 \ g/mol$.
For helium $(He)$,$M_2 = 4 \ g/mol$.
Taking the ratio: $\frac{r_{H_2}}{r_{He}} = \sqrt{\frac{M_{He}}{M_{H_2}}} = \sqrt{\frac{4}{2}} = \sqrt{2} \approx 1.414$.
Therefore,the rate of diffusion of hydrogen is approximately $1.4$ times that of helium.

(C) For a saturated vapor,the vapor pressure is a characteristic property of the substance at a given temperature.
It depends only on the temperature and not on the volume of the container.
When the volume is reduced,some of the vapor condenses into liquid to maintain the equilibrium pressure.
Therefore,the vapor pressure remains constant.

(C) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure. Since the mixture is given by volume percentage,the mole fraction is equal to the volume fraction.
$P_{N_2} = 760 \times \frac{65}{100} = 494 \, mm \, Hg$
$P_{O_2} = 760 \times \frac{15}{100} = 114 \, mm \, Hg$
$P_{CO_2} = 760 \times \frac{20}{100} = 152 \, mm \, Hg$

(D) Let the mass of both ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $m \, g$.
The molar mass of $C_2H_6 = (2 \times 12) + (6 \times 1) = 30 \, g/mol$.
The molar mass of $H_2 = 2 \times 1 = 2 \, g/mol$.
Number of moles of $C_2H_6$ $(n_1)$ = $m / 30$.
Number of moles of $H_2$ $(n_2)$ = $m / 2$.
Total moles $(n_{total})$ = $n_1 + n_2 = (m/30) + (m/2) = (m + 15m) / 30 = 16m / 30 = 8m / 15$.
The partial pressure fraction of a gas is equal to its mole fraction $(x_i = n_i / n_{total})$.
Mole fraction of $H_2$ $(x_{H_2})$ = $(m/2) / (8m/15) = (m/2) \times (15/8m) = 15 / 16$.
Therefore,the fraction of total pressure exerted by hydrogen is $15/16$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r_1/r_2 = \sqrt{M_2/M_1}$.
Given that the rate of the gas $r_g = (1/5) \times r_{H_2}$,we have $r_{H_2}/r_g = 5$.
Substituting this into the formula: $\sqrt{M_g/M_{H_2}} = 5$.
Squaring both sides: $M_g/M_{H_2} = 25$.
Since the molar mass of hydrogen $M_{H_2} = 2 \ g/mol$,we get $M_g = 25 \times 2 = 50 \ g/mol$.

(D) According to Charles's Law,for a fixed mass of gas at constant pressure,the volume is directly proportional to its absolute temperature $(V \propto T)$.
Initial temperature $T_1 = 20 + 273.15 = 293.15 \, K$.
Final temperature $T_2 = 40 + 273.15 = 313.15 \, K$.
Since $T_2 > T_1$,the volume $V$ must increase as $V_2 = V_1 \times (T_2 / T_1)$.
Therefore,the volume increases.

(C) According to Dalton's Law of partial pressures,the total pressure is the sum of the partial pressure of the dry gas and the vapor pressure of water: $P_{\text{total}} = P_{O_2} + P_{H_2O}$.
Given: $P_{\text{total}} = 751\,mm\,Hg$ and $P_{H_2O} = 21\,mm\,Hg$.
Therefore,$P_{O_2} = P_{\text{total}} - P_{H_2O} = 751\,mm\,Hg - 21\,mm\,Hg = 730\,mm\,Hg$.
To convert to $atm$: $P_{O_2} = \frac{730\,mm\,Hg}{760\,mm\,Hg/atm} \approx 0.96\,atm$.

(B) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases present in the mixture. $P_{total} = P_1 + P_2 + ... + P_n$.

(C) Let the weight of both $SO_2$ and $O_2$ be $w \ g$.
Molar mass of $SO_2 = 32 + (2 \times 16) = 64 \ g/mol$.
Molar mass of $O_2 = 2 \times 16 = 32 \ g/mol$.
Number of moles of $SO_2$ $(n_{SO_2})$ = $w/64$.
Number of moles of $O_2$ $(n_{O_2})$ = $w/32$.
Total moles $(n_{total})$ = $w/64 + w/32 = (w + 2w)/64 = 3w/64$.
The partial pressure of a gas is given by its mole fraction multiplied by the total pressure $(P_{O_2} = X_{O_2} \times P_{total})$.
Mole fraction of $O_2$ $(X_{O_2})$ = $n_{O_2} / n_{total} = (w/32) / (3w/64) = (w/32) \times (64/3w) = 2/3$.
Therefore,the ratio of the partial pressure of $O_2$ to the total pressure is $2/3$.

(A) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
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 $32 + (2 \times 16) = 64 \ g/mol$.
The molar mass of $O_2$ is $2 \times 16 = 32 \ g/mol$.
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 $1 : \sqrt{2}$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the rate of diffusion of helium $(He)$ to methane $(CH_4)$ is given by:
$\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{He}}}$
Given molar masses: $M_{He} = 4 \ g/mol$ and $M_{CH_4} = 16 \ g/mol$.
Substituting the values:
$\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.

(A) The total number of moles $n_{total} = n_A + n_B = 5 + 10 = 15 \ mol$.
According to Dalton's Law of partial pressures,$P_i = x_i \times P_{total}$,where $x_i$ is the mole fraction.
$x_A = \frac{5}{15} = \frac{1}{3}$ and $x_B = \frac{10}{15} = \frac{2}{3}$.
$P_A = x_A \times P_{total} = \frac{1}{3} \times 18 = 6 \ atm$.
$P_B = x_B \times P_{total} = \frac{2}{3} \times 18 = 12 \ atm$.

(B) Using the Clausius-Clapeyron equation:
$\log \frac{P_2}{P_1} = \frac{\Delta H_{vap}}{2.303R} \left[ \frac{T_2 - T_1}{T_1 \times T_2} \right]$
Given: $P_1 = 760 \, mm$,$T_1 = 373 \, K$,$P_2 = 23 \, mm$,$\Delta H_{vap} = 40656 \, J/mol$,$R = 8.314 \, J/mol \cdot K$.
Substituting the values:
$\log \frac{760}{23} = \frac{40656}{2.303 \times 8.314} \left[ \frac{373 - T_2}{373 \times T_2} \right]$
$1.519 = 2123.6 \left[ \frac{373 - T_2}{373 \times T_2} \right]$
Solving for $T_2$,we get $T_2 \approx 294.4 \, K$.

(B) The temperature at which the vapor pressure of a liquid becomes equal to the atmospheric pressure is defined as the boiling point of the liquid.

(B) In a pressure cooker,the pressure inside the vessel increases significantly.
According to the relationship between pressure and boiling point,an increase in external pressure leads to an increase in the boiling point of the liquid.
As the boiling point of water rises (above $100 \ ^\circ C$),the food is cooked at a higher temperature,which significantly reduces the cooking time.

(C) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.
The molar mass of $NH_3$ is $17 \ g/mol$ and the molar mass of $CO_2$ is $44 \ g/mol$.
Since $M_{NH_3} < M_{CO_2}$,$NH_3$ will diffuse faster than $CO_2$.
The ratio of their rates of diffusion is given by:
$\frac{r_{NH_3}}{r_{CO_2}} = \sqrt{\frac{M_{CO_2}}{M_{NH_3}}} = \sqrt{\frac{44}{17}} \approx 1.61$.

(C) For gases at different pressures,the rate of effusion $r$ is given by Graham's Law modified for pressure:
$r = \frac{P \times A}{\sqrt{2 \pi M R T}} \propto \frac{P}{\sqrt{M}}$
Thus,$\frac{r_1}{r_2} = \frac{P_1}{P_2} \sqrt{\frac{M_2}{M_1}}$
Given: $P_{N_2} = 0.8 \, atm$,$t_{N_2} = 38 \, s$,$M_{N_2} = 28 \, g/mol$,$P_X = 1.6 \, atm$,$t_X = 57 \, s$.
Since $r = \frac{n}{t}$,we have $\frac{r_{N_2}}{r_X} = \frac{t_X}{t_{N_2}} = \frac{57}{38} = 1.5$.
Substituting into the formula: $1.5 = \frac{0.8}{1.6} \sqrt{\frac{M_X}{28}}$
$1.5 = 0.5 \times \sqrt{\frac{M_X}{28}}$
$3 = \sqrt{\frac{M_X}{28}}$
$9 = \frac{M_X}{28}$
$M_X = 9 \times 28 = 252 \, g/mol$.

(C) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$.
Rearranging for density $(d = \frac{m}{V})$,we get $d = \frac{PM}{RT}$.
Since $M$ (molar mass) and $R$ (gas constant) are constant for a given gas,$d \propto \frac{P}{T}$.
To maximize density $(d)$,we need the highest pressure $(P)$ and the lowest temperature $(T)$.
Comparing the given conditions:
$A$: $T = 373 \, K$
$B$: $T = 546 \, K$
$C$: $T = 273 \, K$
$D$: $T = 298 \, K$
The lowest temperature is $0 \, ^\circ C$ $(273 \, K)$ at $1 \, atm$ pressure.
Therefore,the density is maximum at $0 \, ^\circ C$ and $1 \, atm$.

(B) Pressure is defined as force per unit area $(P = F/A)$.
Energy per unit volume is given by $\frac{\text{Energy}}{\text{Volume}} = \frac{\text{Force} \times \text{Distance}}{\text{Area} \times \text{Distance}} = \frac{\text{Force}}{\text{Area}} = \text{Pressure}$.
Therefore,the unit of pressure is equivalent to energy per unit volume.

(A) Absolute zero ($0 \, K$ or $-273.15 \, ^oC$) is the theoretical temperature at which all molecular motion is assumed to cease.

(D) The density of the gas is $2.86 \, g \, L^{-1}$,which means $1 \, L$ of the gas has a mass of $2.86 \, g$.
At $STP$,$1 \, \text{mole}$ of any ideal gas occupies $22.4 \, L$.
Therefore,the molar mass of the gas = $\text{Density} \times 22.4 \, L \, \text{mol}^{-1}$.
Molar mass = $2.86 \, g \, L^{-1} \times 22.4 \, L \, \text{mol}^{-1} = 64.064 \, g \, \text{mol}^{-1} \approx 64 \, g \, \text{mol}^{-1}$.
The molar mass of $SO_2$ is $32 + (2 \times 16) = 64 \, g \, \text{mol}^{-1}$.
Thus,the gas is $SO_2$.

(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 $M_{H_2} = 2 \ g/mol$ and $M_{gas} = 98 \ g/mol$.
The relative rate of diffusion of the gas with respect to hydrogen is $\frac{r_{gas}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{gas}}}$.
Substituting the values: $\sqrt{\frac{2}{98}} = \sqrt{\frac{1}{49}} = \frac{1}{7}$.

(A) According to Graham's Law of Diffusion,the rate of diffusion of a gas is inversely proportional to the square root of its molar mass $(r \propto 1/\sqrt{M})$.
Therefore,gases with the same molar mass will diffuse at the same rate.
The molar mass of $CO_2 = 12 + (2 \times 16) = 44 \ g/mol$.
The molar mass of $N_2O = (2 \times 14) + 16 = 44 \ g/mol$.
Since both $CO_2$ and $N_2O$ have the same molar mass of $44 \ g/mol$,they will diffuse at the same rate.

(A) According to Graham's Law of Diffusion,for gases diffusing in the same time $t$ under identical conditions of temperature and pressure,the rate of diffusion $r$ is proportional to the volume $V$ diffused $(r = V/t)$.
Thus,$\frac{r_A}{r_B} = \frac{V_A}{V_B} = \sqrt{\frac{M_B}{M_A}}$.
Given $V_A = 50 \ mL$,$V_B = 40 \ mL$,and $M_A = 64$.
Substituting the values: $\frac{50}{40} = \sqrt{\frac{M_B}{64}}$.
Squaring both sides: $(\frac{5}{4})^2 = \frac{M_B}{64}$.
$\frac{25}{16} = \frac{M_B}{64}$.
$M_B = \frac{25}{16} \times 64 = 25 \times 4 = 100$.

(A) According to Graham's Law of Diffusion,$\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}}$.
Given that $r_A = 5r_B$,we have $\frac{r_A}{r_B} = 5$.
Substituting this into the equation: $5 = \sqrt{\frac{d_B}{d_A}}$.
Squaring both sides: $25 = \frac{d_B}{d_A}$.
Therefore,the ratio of densities $\frac{d_A}{d_B} = \frac{1}{25} = 0.04$.

(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the density $(d)$: $r \propto \frac{1}{\sqrt{d}}$.
Given the ratio of densities $\frac{d_1}{d_2} = \frac{1}{16}$.
The ratio of rates of diffusion is $\frac{r_1}{r_2} = \sqrt{\frac{d_2}{d_1}}$.
Substituting the values: $\frac{r_1}{r_2} = \sqrt{\frac{16}{1}} = 4$.
Therefore,the ratio of their rates of diffusion is $4:1$.