Osmosis and Osmotic pressure of the solution Questions in English

(B) semi-permeable membrane is a type of biological or synthetic,polymeric membrane that will allow certain molecules or ions to pass through it by diffusion or occasionally by more specialized processes like facilitated diffusion,passive transport,or active transport. In the context of osmosis,it specifically allows the passage of solvent molecules (usually water) while blocking the passage of solute molecules.

(B) For isotonic solutions,the molar concentrations are equal: $C_1 = C_2$.
Concentration of urea $(C_1)$:
$C_1 = \frac{\text{mass}}{\text{molar mass} \times \text{volume in L}} = \frac{8.6 \ g}{60 \ g/mol \times 1 \ L} = \frac{8.6}{60} \ mol/L$.
Concentration of organic solute $(C_2)$:
$0.5\% \ (w/v)$ means $0.5 \ g$ in $100 \ mL$,which is $0.5 \ g$ in $0.1 \ L$.
$C_2 = \frac{0.5 \ g}{M \times 0.1 \ L} = \frac{5}{M} \ mol/L$,where $M$ is the molar mass.
Equating $C_1 = C_2$:
$\frac{8.6}{60} = \frac{5}{M}$
$M = \frac{5 \times 60}{8.6} = \frac{300}{8.6} \approx 34.88 \ g/mol$.

(A) $12\%$ solution means $12\,g$ of cane sugar is present in $100\,mL$ of the solution.
Number of moles of sugar $(n)$ = $\frac{12}{342} \approx 0.0351\,mol$.
Volume of solution $(V)$ = $100\,mL = 0.1\,L$.
Temperature $(T)$ = $17 + 273 = 290\,K$.
Using the osmotic pressure formula $\pi = \frac{nRT}{V}$:
$\pi = \frac{0.0351 \times 0.0821 \times 290}{0.1} = 8.35\,atm$.

(D) The formula for osmotic pressure is $\pi = CRT$.
Given: $\pi = 2.5 \ atm$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 24 + 273 = 297 \ K$.
Substituting the values: $2.5 = C \times 0.0821 \times 297$.
$C = \frac{2.5}{0.0821 \times 297} \approx 0.1025 \ mol \ L^{-1}$.

(D) For isotonic solutions,the osmotic pressures are equal: $\pi_1 = \pi_2$.
Since the solvent and temperature are the same,the molar concentrations $(C)$ must be equal: $C_1 = C_2$.
The molar concentration is given by $\frac{w \, (g/L)}{Molar \, mass \, (g/mol)}$.
For the substance: $C_1 = \frac{5.25 \, g/100 \, mL}{M_1} = \frac{52.5}{M_1} \, mol/L$.
For urea: $C_2 = \frac{1.5 \, g/100 \, mL}{60 \, g/mol} = \frac{15}{60} \, mol/L = 0.25 \, mol/L$.
Equating the two: $\frac{52.5}{M_1} = 0.25$.
$M_1 = \frac{52.5}{0.25} = 210 \, g \, mol^{-1}$.

(C) Osmosis is the movement of solvent molecules from a region of lower solute concentration (hypotonic) to a region of higher solute concentration (hypertonic) through a semi-permeable membrane.
If solution $A$ is isotonic to solution $B$,the osmotic pressure of both solutions is equal.
Therefore,there is no net movement of solvent molecules between them.
Thus,osmosis does not occur when $A$ is isotonic to $B$.

(B) Given,$\pi_1 = 4.92\,atm$,$\pi_2 = 1.5\,atm$.
For glucose $(C_6H_{12}O_6)$,molar mass $M = 180\,g/mol$.
Concentration $C_1 = \frac{36\,g}{180\,g/mol \times 1\,L} = 0.2\,M$.
Since osmotic pressure $\pi = CRT$,at constant temperature $T$,$\pi \propto C$.
Therefore,$\frac{\pi_1}{\pi_2} = \frac{C_1}{C_2}$.
$\frac{4.92}{1.5} = \frac{0.2}{C_2}$.
$C_2 = \frac{0.2 \times 1.5}{4.92} = \frac{0.3}{4.92} \approx 0.061\,mol/L$.

(C) Osmotic pressure $(\Pi)$ is given by the formula $\Pi = CRT = \frac{n}{V}RT = \frac{w}{M \times V}RT$.
Since $w$ (mass),$V$ (volume),$R$ (gas constant),and $T$ (temperature) are constant for all three solutions,the osmotic pressure is inversely proportional to the molar mass $(M)$ of the solute: $\Pi \propto \frac{1}{M}$.
The molar masses are: Urea $(NH_2CONH_2)$ = $60 \ g/mol$,Glucose $(C_6H_{12}O_6)$ = $180 \ g/mol$,and Sucrose $(C_{12}H_{22}O_{11})$ = $342 \ g/mol$.
Since $60 < 180 < 342$,the osmotic pressures follow the order $P_2 > P_1 > P_3$.

(A) solution that has the same osmotic pressure as the fluid inside the red blood cells is known as an $Isotonic$ solution. In an $Isotonic$ solution,there is no net movement of water across the cell membrane,allowing the red blood cells to maintain their normal shape and size.

(C) For isotonic solutions,the molar concentrations are equal at the same temperature.
$Molar \ mass \ of \ sucrose \ (C_{12}H_{22}O_{11}) = 342 \ g/mol$.
$Molar \ concentration \ of \ sucrose = \frac{34.2 \ g/L}{342 \ g/mol} = 0.1 \ mol/L$.
For the glucose solution to be isotonic,its molar concentration must also be $0.1 \ mol/L$.
$Molar \ mass \ of \ glucose \ (C_6H_{12}O_6) = 180 \ g/mol$.
$Concentration \ of \ glucose = 0.1 \ mol/L \times 180 \ g/mol = 18 \ g/L$.

(C) The formula for osmotic pressure is $\pi = CRT = \frac{n}{V}RT = \frac{w}{M \times V}RT$.
Given: $w = 5\,g$ (in $100\,mL$ of solution),$M = 342\,g/mol$,$T = 150 + 273 = 423\,K$,$V = 100\,mL = 0.1\,L$,and $R = 0.0821\,L\,atm\,K^{-1}mol^{-1}$.
Substituting the values: $\pi = \frac{5 \times 0.0821 \times 423}{342 \times 0.1} = \frac{173.631}{34.2} \approx 5.07\,atm$.

(D) Two solutions are isotonic if they have the same osmotic pressure $(\pi = iCRT)$. Since the temperature $(T)$ and concentration $(C)$ are the same,the van't Hoff factor $(i)$ must be equal for the solutions to be isotonic.
For $0.1 \ M \ Ca(NO_3)_2$: $i = 3$ $(Ca^{2+} + 2NO_3^-)$,so effective concentration = $0.1 \times 3 = 0.3 \ M$.
For $0.1 \ M \ Na_2SO_4$: $i = 3$ $(2Na^{+} + SO_4^{2-})$,so effective concentration = $0.1 \times 3 = 0.3 \ M$.
Since both have the same effective concentration of particles,they are isotonic.

(A) Osmotic pressure $(\pi)$ is a colligative property given by $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since all solutions are decimolar $(C = 0.1 \ M)$,the osmotic pressure depends on the van't Hoff factor $(i = 1 + (n-1)\alpha)$.
For $90\%$ dissociation,$\alpha = 0.9$.
$(A)$ $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,$n = 5$,$i = 1 + (5-1)(0.9) = 4.6$.
$(B)$ $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,$n = 3$,$i = 1 + (3-1)(0.9) = 2.8$.
$(C)$ $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,$n = 3$,$i = 1 + (3-1)(0.9) = 2.8$.
$(D)$ Mixing $(B)$ and $(C)$ results in a solution with lower concentration of individual ions compared to $(A)$.
Thus,$Al_2(SO_4)_3$ provides the maximum number of ions,resulting in the highest osmotic pressure.

(D) The formula for osmotic pressure is $\pi V = nRT$,where $n = \frac{w}{M}$.
Given:
$\pi = 2.57 \times 10^{-3} \ bar$
$V = 200 \ mL = 0.2 \ L$
$w = 1.26 \ g$
$T = 300 \ K$
$R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1}$
Substituting the values into the equation $\pi V = \frac{w}{M} RT$:
$2.57 \times 10^{-3} \times 0.2 = \frac{1.26}{M} \times 0.083 \times 300$
$M = \frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.2}$
$M = \frac{31.374}{0.000514} \approx 61039 \ g \ mol^{-1}$.

(D) Osmosis is the phenomenon of the flow of solvent molecules through a semipermeable membrane.
In reality,solvent molecules move in both directions across the membrane.
However,the net flow occurs from the region of lower solute concentration (higher solvent concentration) to the region of higher solute concentration (lower solvent concentration).
Since no membrane is perfectly semipermeable,the flow of water occurs from both sides but at unequal rates,resulting in a net flow.

(A) For isotonic solutions,the osmotic pressure $(\pi)$ is equal,which implies that their molar concentrations $(C)$ are equal at the same temperature.
Given that the solutions are $5.25\%$ and $1.5\%$ by mass,and assuming the density is $1.0\ g\ cm^{-3}$,the concentration in $g\ L^{-1}$ is the same as the percentage mass multiplied by $10$.
Concentration of substance $(C_1) = \frac{5.25 \times 10}{M} = \frac{52.5}{M} \ mol\ L^{-1}$.
Concentration of urea $(C_2) = \frac{1.5 \times 10}{60} = \frac{15}{60} = 0.25 \ mol\ L^{-1}$.
Since the solutions are isotonic,$C_1 = C_2$.
$\frac{52.5}{M} = 0.25$.
$M = \frac{52.5}{0.25} = 210 \ g\ mol^{-1}$.

(A) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For $C_2H_5OH$ (non-electrolyte): $i = 1$,$\pi = 1 \times 0.500 \times RT = 0.500 RT$.
For $Mg_3(PO_4)_2$ (dissociates into $3Mg^{2+} + 2PO_4^{3-}$,so $i = 5$): $\pi = 5 \times 0.100 \times RT = 0.500 RT$.
For $KBr$ (dissociates into $K^+ + Br^-$,so $i = 2$): $\pi = 2 \times 0.250 \times RT = 0.500 RT$.
For $Na_3PO_4$ (dissociates into $3Na^+ + PO_4^{3-}$,so $i = 4$): $\pi = 4 \times 0.125 \times RT = 0.500 RT$.
Since the osmotic pressure of all the given solutions is equal,they are isotonic.

(C) Two solutions are isotonic if their osmotic pressures are equal,i.e.,$\pi_1 = \pi_2$.
For a non-electrolyte like sugar,the osmotic pressure is $\pi = C \times R \times T$.
For an electrolyte like $KCl$,the osmotic pressure is $\pi = i \times C \times R \times T$,where $i$ is the van't Hoff factor.
Since $KCl$ dissociates as $KCl \rightarrow K^+ + Cl^-$,the number of ions produced is $2$,so $i = 2$.
Given the normality $(N)$ of $KCl$ is $0.5 \ N$. Since the n-factor for $KCl$ is $1$,the molarity $(M)$ is equal to the normality $(N)$,so $M = 0.5 \ M$.
Equating the osmotic pressures: $i_{KCl} \times M_{KCl} = M_{sugar}$.
$2 \times 0.5 = M_{sugar}$.
$M_{sugar} = 1 \ M$.
Since the n-factor for sugar is $1$,the normality of the sugar solution is $1 \ N$.

(D) The osmotic pressure $(\pi)$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature. Since $C$,$R$,and $T$ are constant for all options,$\pi$ is directly proportional to $i$.
For $CH_3COOH$: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$. $i = 1 + \alpha = 1 + 0.7 = 1.7$.
For $KCl$: $KCl \rightarrow K^+ + Cl^-$. $i = 2$.
For $Na_2SO_4$: $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$. $i = 3$.
For $K_2Zn[Fe(CN)_6]$: $K_2Zn[Fe(CN)_6] \rightarrow 2K^+ + Zn[Fe(CN)_6]^{2-}$. $i = 3$.
Wait,let's re-evaluate $K_2Zn[Fe(CN)_6]$. It dissociates into $2K^+$ and $Zn[Fe(CN)_6]^{2-}$,so $i = 3$.
Comparing $i$ values: $1.7, 2, 3, 3$.
Actually,$Na_2SO_4$ and $K_2Zn[Fe(CN)_6]$ both have $i = 3$. However,in standard chemistry problems of this type,$K_2Zn[Fe(CN)_6]$ is often treated as a complex salt. Let's re-check the dissociation: $K_2Zn[Fe(CN)_6] \rightarrow 2K^+ + Zn^{2+} + [Fe(CN)_6]^{4-}$,which would give $i = 4$. Given the options,$K_2Zn[Fe(CN)_6]$ provides the highest number of ions.

(D) Given: $w/v\ \% = 2.22\ \%$,which means $2.22\ g$ of $CaCl_2$ in $100\ mL$ of solution.
$T = 27\ ^oC = 300\ K$.
$w_B = 2.22\ g$,$v = 0.1\ L$.
$M_B$ of $CaCl_2 = 40 + 2 \times 35.5 = 111\ g/mol$.
Number of moles $n = \frac{2.22}{111} = 0.02\ mol$.
For $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$,the van't Hoff factor $i = 3$.
Osmotic pressure $\pi = i \times \frac{n}{v} \times R \times T$.
$\pi = 3 \times \frac{0.02}{0.1} \times 0.08 \times 300$.
$\pi = 3 \times 0.2 \times 0.08 \times 300 = 14.4\ atm$.

(D) The osmotic pressure $(\pi)$ is given by the formula: $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molarity $(M)$,$R$ is the universal gas constant,and $T$ is the temperature.
Comparing this with the equation of a straight line $y = mx + c$ (where $y = \pi$ and $x = M$):
$\pi = (iRT) \times M$
Here,the slope $(m)$ is equal to $iRT$.
Since $NaCl$ is an electrolyte,its van't Hoff factor $i > 1$.
Therefore,the slope of the line represents $iRT$,which is not explicitly listed as an option (as $iRT \neq RT$ for $NaCl$).
Thus,the correct option is $D$.

(B) Two solutions are isotonic if they have the same molar concentration $(C_1 = C_2)$.
For urea solution: $C_{\text{urea}} = \frac{\text{mass}}{\text{molar mass} \times \text{volume}} = \frac{6 \ g}{60 \ g/mol \times 1 \ L} = 0.1 \ mol/L$.
Since the solutions are isotonic,the concentration of cane sugar must also be $0.1 \ mol/L$.
Concentration of sugar in $g/L = \text{Molarity} \times \text{Molar mass} = 0.1 \ mol/L \times 342 \ g/mol = 34.2 \ g/L$.

(C) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration $(C_1 = C_2)$.
For a $5\% \, w/v$ solution of cane sugar $(C_{12}H_{22}O_{11})$,the mass of solute is $5 \ g$ in $100 \ mL$ of solution.
Molar mass of cane sugar $= 342 \ g/mol$.
Molarity $(C_1)$ $= \frac{5 \ g}{342 \ g/mol} \times \frac{1000 \ mL}{100 \ mL} = \frac{50}{342} \ M$.
For a $1\% \, w/v$ solution of substance $X$,the mass of solute is $1 \ g$ in $100 \ mL$ of solution.
Let the molar mass of $X$ be $M$.
Molarity $(C_2)$ $= \frac{1 \ g}{M \ g/mol} \times \frac{1000 \ mL}{100 \ mL} = \frac{10}{M} \ M$.
Since the solutions are isotonic,$C_1 = C_2$:
$\frac{50}{342} = \frac{10}{M}$
$M = \frac{10 \times 342}{50} = \frac{342}{5} = 68.4 \ g/mol$.

(B) semipermeable membrane is a type of biological or synthetic polymeric membrane that allows certain molecules or ions to pass through it by diffusion.
In the context of osmotic pressure,it specifically allows the passage of $solvent$ molecules while restricting the passage of $solute$ molecules.
Examples include animal bladders or cellophane membranes.

(D) Two solutions are isotonic if they have the same osmotic pressure,i.e.,$\pi_1 = \pi_2$.
Since $\pi = CRT$,for isotonic solutions at the same temperature,$C_1 = C_2$.
For the first substance: Concentration $C_1 = \frac{1.05 \ g}{M \times 0.1 \ L} = \frac{10.5}{M} \ mol/L$.
For the $3\%$ glucose solution: $3 \ g$ of glucose in $100 \ mL$ solution. Molar mass of glucose $(C_6H_{12}O_6)$ = $180 \ g/mol$.
Concentration $C_2 = \frac{3 \ g}{180 \ g/mol \times 0.1 \ L} = \frac{30}{180} = \frac{1}{6} \ mol/L$.
Equating $C_1 = C_2$: $\frac{10.5}{M} = \frac{1}{6}$.
$M = 10.5 \times 6 = 63 \ g/mol$.

(C) Two solutions are isotonic if they have the same osmotic pressure,i.e.,$\pi_1 = \pi_2$.
For solutions with the same temperature and volume,this implies $C_1 = C_2$ or $\frac{n_1}{V} = \frac{n_2}{V}$.
Given: Urea solution is $1.5 \% \, (w/V)$,which means $1.5 \, g$ of urea in $100 \, mL$ of solution.
Molar mass of urea $(NH_2CONH_2) = 60 \, g/mol$.
Molar mass of glucose $(C_6H_{12}O_6) = 180 \, g/mol$.
Using the relation $\frac{w_1}{M_1} = \frac{w_2}{M_2}$:
$\frac{1.5}{60} = \frac{w_2}{180}$.
$w_2 = \frac{1.5 \times 180}{60} = 1.5 \times 3 = 4.5 \, g$.
Thus,a $4.5 \% \, (w/V)$ glucose solution is isotonic with $1.5 \% \, (w/V)$ urea solution.

(B) The osmotic pressure $\pi$ is given by $\pi = h \cdot d \cdot g$,where $h$ is the height of the solution column,$d$ is the density,and $g$ is the acceleration due to gravity.
Given: $h = 11 \, cm = 0.11 \, m$,$d = 0.9 \, g/mL = 900 \, kg/m^3$,$C = 10 \, g/L = 10/M \, mol/L = 10/M \times 10^3 \, mol/m^3$,$T = 27 + 273 = 300 \, K$,$R = 8.314 \, J/(mol \cdot K)$.
Using $\pi = CRT$:
$h \cdot d \cdot g = (C/M) \cdot R \cdot T$
$0.11 \times 900 \times 9.8 = (10/M) \times 10^3 \times 8.314 \times 300$
$970.2 = (24942000 / M)$
$M = 24942000 / 970.2 \approx 25708 \, g/mol$.
Using the standard approximation $\pi = h \cdot d \cdot g$ and converting units to $atm$:
$\pi = (11 \, cm \times 0.9 \, g/mL) / (13.6 \, g/mL \times 76 \, cm) \approx 0.00958 \, atm$.
Using $\pi = CRT$ with $C = 10/M \, g/L$:
$0.00958 = (10/M) \times 0.0821 \times 300$
$M = (10 \times 0.0821 \times 300) / 0.00958 \approx 25.2 \times 10^3 \, g/mol$.

(B) Two solutions are isotonic if they have the same molar concentration $(C_1 = C_2)$.
For urea: Concentration $= \frac{10 \, g \, dm^{-3}}{60 \, g \, mol^{-1}} = \frac{1}{6} \, mol \, dm^{-3}$.
For the non-volatile solute: $A$ $5 \%$ solution means $5 \, g$ of solute in $100 \, mL$ of solution,which is equivalent to $50 \, g \, dm^{-3}$.
Let the molecular mass of the non-volatile solute be $M$.
Concentration of non-volatile solute $= \frac{50 \, g \, dm^{-3}}{M \, g \, mol^{-1}} = \frac{50}{M} \, mol \, dm^{-3}$.
Since the solutions are isotonic,$\frac{1}{6} = \frac{50}{M}$.
Therefore,$M = 50 \times 6 = 300 \, g \, mol^{-1}$.

(B) The osmotic pressure is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.
For $NaCl$ (strong electrolyte),$i = 2$,so $\pi_a = 2 \times 2 \times RT = 4RT$.
For glucose (non-electrolyte),$i = 1$,so $\pi_b = 1 \times 1 \times RT = 1RT$.
For urea (non-electrolyte),$i = 1$,so $\pi_c = 1 \times 2 \times RT = 2RT$.
Comparing the values,the $1 \ M$ glucose solution has the minimum osmotic pressure.

(A) The osmotic pressure $(\pi)$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For sucrose (a non-electrolyte),the van't Hoff factor $i = 1$. Thus,$\pi_{\text{sucrose}} = 1 \times C \times R \times T = 0.24 \ atm$.
For $KI$ (an electrolyte),$\pi_{KI} = i_{KI} \times C \times R \times T = 0.432 \ atm$.
Dividing the two equations:
$\frac{\pi_{KI}}{\pi_{\text{sucrose}}} = \frac{i_{KI} \times C \times R \times T}{1 \times C \times R \times T} = \frac{0.432}{0.24}$.
$i_{KI} = 1.80$.

(B) The osmotic pressure is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
For $KCl$,the dissociation is $KCl \rightarrow K^{+} + Cl^{-}$,so $n = 2$.
Given the degree of dissociation $\alpha = 0.8$,the van't Hoff factor is $i = 1 + \alpha(n - 1) = 1 + 0.8(2 - 1) = 1.8$.
$A$ $1\% \ (wt/vol)$ solution means $1 \ g$ of $KCl$ in $100 \ mL$ of solution,which is $10 \ g/L$.
The molar mass of $KCl$ is $39.1 + 35.5 = 74.6 \ g/mol$.
Concentration $C = \frac{10 \ g/L}{74.6 \ g/mol} \approx 0.134 \ mol/L$.
Temperature $T = 27 + 273 = 300 \ K$.
$\pi = 1.8 \times 0.134 \times 0.0821 \times 300 \approx 5.94 \ atm$.

(D) semipermeable membrane allows only the passage of solvent molecules (water) and restricts the movement of solute particles.
Osmosis involves the net movement of solvent from a region of lower solute concentration to a region of higher solute concentration.
Since the solutes cannot cross the membrane,they cannot come into contact to react and form the brown-colored product.
Therefore,no brown color formation occurs in either side.

(C) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the total molarity of the solution.
First,calculate the moles of urea: $n_{\text{urea}} = \frac{0.6 \ g}{60 \ g \ mol^{-1}} = 0.01 \ mol$.
Next,calculate the moles of glucose: $n_{\text{glucose}} = \frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$.
Total moles of solute $= 0.01 + 0.01 = 0.02 \ mol$.
The volume of the solution is $100 \ mL = 0.1 \ L$.
Total molarity $C = \frac{0.02 \ mol}{0.1 \ L} = 0.2 \ M$.
Temperature $T = 27 \ ^oC = 27 + 273 = 300 \ K$.
Now,calculate osmotic pressure: $\pi = 0.2 \ mol \ L^{-1} \times 0.08206 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 0.2 \times 0.08206 \times 300 = 4.9236 \ atm$.

(C) Osmosis is the process in which solvent molecules (water) move from a region of lower solute concentration to a region of higher solute concentration through a semipermeable membrane $(SPM)$.
Since solution $A$ $(2\% \ NaCl)$ has a lower concentration than solution $B$ $(10\% \ NaCl)$,water will flow from $A$ to $B$ to equalize the concentrations.

(A) Osmotic pressure $(O.P.)$ is a colligative property,which depends on the number of particles in the solution.
For equimolar solutions,the osmotic pressure is proportional to the van't Hoff factor $(i)$.
$BaCl_2$ dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,so $i = 3$.
$NaCl$ dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
Glucose is a non-electrolyte,so $i = 1$.
Thus,the order of the number of particles is $BaCl_2 (3) > NaCl (2) >$ Glucose $(1)$.
Therefore,the order of osmotic pressure is $BaCl_2 > NaCl >$ Glucose.

(D) The process of obtaining fresh water from sea water is known as $Reverse \ Osmosis$.
In this process,a pressure greater than the osmotic pressure is applied to the saline water (sea water) across a semi-permeable membrane.
This forces the water molecules to move from the concentrated solution (sea water) to the dilute solution (fresh water),effectively separating the salt and other impurities.

(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{i-1}{n-1}$,where $n$ is the number of ions produced per formula unit of $NaCl$ $(n=2)$.
Given $\alpha = 0.9$,we have $0.9 = \frac{i-1}{2-1}$,which gives the van't Hoff factor $i = 1.9$.
The osmotic pressure $\pi$ is calculated using the formula $\pi = i \times C \times R \times T$.
Here,$C = 0.002 \, M$,$R = 0.082 \, L \cdot bar \cdot K^{-1} \cdot mol^{-1}$,and $T = 27 + 273 = 300 \, K$.
Substituting the values: $\pi = 1.9 \times 0.002 \times 0.082 \times 300 = 0.09348 \, bar \approx 0.094 \, bar$.

(C) The formula for osmotic pressure is $\pi = \frac{wRT}{MV}$,where $w$ is the mass of solute,$R$ is the gas constant,$T$ is the temperature in Kelvin,$M$ is the molar mass of solute,and $V$ is the volume of solution in liters.
Given: $w = 5 \ g$,$V = 100 \ mL = 0.1 \ L$,$T = 150 + 273 = 423 \ K$,$M = 342 \ g/mol$ (for cane sugar $C_{12}H_{22}O_{11}$),and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $\pi = \frac{5 \times 0.0821 \times 423}{342 \times 0.1} = \frac{173.63}{34.2} \approx 5.077 \ atm$.

(A) Osmosis is the spontaneous flow of solvent molecules through a semipermeable membrane from a region of pure solvent (or dilute solution) to a region of concentrated solution.
According to the vapour pressure theory,the vapour pressure of a pure solvent is always higher than that of a solution.
Since solvent molecules move from a region of higher chemical potential (higher vapour pressure) to a region of lower chemical potential (lower vapour pressure),the correct direction is from higher vapour pressure to lower vapour pressure.

(A) The formula for osmotic pressure is $\pi = i \times C \times R \times T$.
Given: $\pi = 4.92 \ atm$,$C = 0.1 \ M$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 300 \ K$.
Substituting the values: $4.92 = i \times 0.1 \times 0.0821 \times 300$.
$4.92 = i \times 2.463$.
$i = \frac{4.92}{2.463} \approx 1.998 \approx 2$.
For $MgCl_2$,the dissociation is $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$,so $n = 3$.
The degree of ionization $\alpha$ is given by $\alpha = \frac{i-1}{n-1}$.
$\alpha = \frac{1.998-1}{3-1} = \frac{0.998}{2} = 0.499$.
Percentage ionization $= \alpha \times 100 = 49.9 \% \approx 50 \%$. Given the options,$49 \%$ is the closest value.

(B) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.
For $NaCl$ (strong electrolyte),$i = 2$ (dissociates into $Na^+$ and $Cl^-$). Thus,$\pi_a = 2 \times 2 \ RT = 4 \ RT$.
For glucose (non-electrolyte),$i = 1$. Thus,$\pi_b = 1 \times 1 \ RT = 1 \ RT$.
For urea (non-electrolyte),$i = 1$. Thus,$\pi_c = 1 \times 2 \ RT = 2 \ RT$.
Comparing the values,the glucose solution has the minimum osmotic pressure.

(A) The formula for osmotic pressure is $\pi V = nRT = \frac{W_B}{M_B} RT$.
Rearranging for molecular weight $M_B = \frac{W_B \times R \times T}{\pi \times V}$.
Given: $W_B = 20 \ g$,$V = 500 \ mL = 0.5 \ L$,$T = 15 + 273 = 288 \ K$,$\pi = \frac{600}{760} \ atm$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $M_B = \frac{20 \times 0.0821 \times 288}{(600/760) \times 0.5}$.
$M_B = \frac{20 \times 0.0821 \times 288 \times 760}{600 \times 0.5} \approx 1198 \ g \ mol^{-1}$.

(A) In beaker $X$,the grape remains the same size,indicating an isotonic solution where the net flow of water is zero.
In beaker $Y$,the grape swells,which indicates that water has entered the grape due to osmosis. This happens when the surrounding solution is hypotonic (lower osmotic pressure) compared to the grape's internal solution.
In beaker $Z$,the grape shrinks,which indicates that water has left the grape due to osmosis. This happens when the surrounding solution is hypertonic (higher osmotic pressure) compared to the grape's internal solution.
Therefore,$Y$ is a hypotonic solution and $Z$ is a hypertonic solution.

(D) The osmotic pressure $(\pi)$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
For $0.011 \ M \ AlCl_3$ $(i=4)$: $\pi = 4 \times 0.011 \times R \times 323 = 0.014212 \ R$.
For $0.03 \ M \ NaCl$ $(i=2)$: $\pi = 2 \times 0.03 \times R \times 298 = 0.01788 \ R$.
For $0.012 \ M \ (NH_4)_2SO_4$ $(i=3)$: $\pi = 3 \times 0.012 \times R \times 298 = 0.010728 \ R$.
For $0.03 \ M \ NaCl$ at $50 \ ^oC$ $(i=2)$: $\pi = 2 \times 0.03 \times R \times 323 = 0.01938 \ R$.
Comparing the values,$0.03 \ M \ NaCl$ at $50 \ ^oC$ has the highest osmotic pressure.

(A) Given: Mass of solute $(CH_3COOH)$ = $90 \, g$,Molar mass of $CH_3COOH$ = $60 \, g/mol$,Mass of solvent (benzene) = $110 \, g$,Density of solution $(d)$ = $1.2 \, g/mL$,$i = 0.5$,$T = 300 \, K$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
Total mass of solution = $90 \, g + 110 \, g = 200 \, g$.
Volume of solution $(V)$ = $\frac{\text{mass}}{\text{density}} = \frac{200 \, g}{1.2 \, g/mL} = 166.67 \, mL = 0.16667 \, L$.
Moles of solute $(n)$ = $\frac{90 \, g}{60 \, g/mol} = 1.5 \, mol$.
Molarity $(C)$ = $\frac{n}{V(L)} = \frac{1.5}{0.16667} \approx 9 \, M$.
Using the formula for osmotic pressure: $\pi = iCRT$.
$\pi = 0.5 \times 9 \times 0.0821 \times 300 = 110.835 \, atm \approx 110.8 \, atm$.

(A) The dissociation of Aluminium sulphate is: $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$.
Here,the number of ions produced per formula unit is $n = 5$.
The van't Hoff factor $i$ is calculated as: $i = 1 + (n-1)\alpha = 1 + (5-1) \times 0.9 = 1 + 3.6 = 4.6$.
Given $Normality = 0.03 \ N$ and the $n$-factor for $Al_2(SO_4)_3$ is $6$ (total positive charge),the Molarity $M$ is: $M = \frac{Normality}{n\text{-factor}} = \frac{0.03}{6} = 0.005 \ M$.
The osmotic pressure $\pi$ is given by: $\pi = iCRT$.
Substituting the values: $\pi = 4.6 \times 0.005 \times 0.0821 \times 300 \ K$.
$\pi = 0.566 \ atm$.

(D) The semipermeable membrane $(SPM)$ allows only solvent molecules to pass through it,not the solute ions ($Fe^{3+}$,$Cl^-$,$K^+$,$[Fe(CN)_6]^{4-}$).
Since the solute ions cannot cross the $SPM$,they cannot come into contact with each other to react and form the blue-colored complex (Prussian blue).
Osmosis involves the movement of solvent molecules from a region of lower solute concentration to a region of higher solute concentration,but this does not facilitate the mixing of the solutes themselves.
Therefore,no reaction occurs,and no blue color is formed.

(C) The osmotic pressure $\pi$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For $0.2 \ M$ urea: $\pi_1 = 0.2 \times RT$.
For $0.2 \ M$ glucose: $\pi_2 = 0.2 \times RT$.
When equal volumes $(V)$ of these two solutions are mixed,the new concentration $C_{mix}$ is calculated as:
$C_{mix} = \frac{C_1 V + C_2 V}{V + V} = \frac{0.2 V + 0.2 V}{2V} = 0.2 \ M$.
Therefore,the osmotic pressure of the mixture is $\pi_{mix} = 0.2 \times RT$,which is the same as the osmotic pressure of the individual solutions.

(A) Two solutions are isotonic if they have the same osmotic pressure,i.e.,$\pi_1 = \pi_2$.
Since $\pi = iCRT$,at the same temperature,the condition for isotonic solutions is $i_1 C_1 = i_2 C_2$,where $i$ is the van't Hoff factor and $C$ is the molar concentration.
For $1.5 \ M \ AlCl_3$: $i = 4$ (dissociates into $Al^{3+} + 3Cl^-$),so $i_1 C_1 = 4 \times 1.5 = 6.0$.
For $1 \ M \ Na_2SO_4$: $i = 3$ (dissociates into $2Na^+ + SO_4^{2-}$),so $i_2 C_2 = 3 \times 1 = 3.0$.
Checking option $C$: $1.5 \ M \ AlCl_3$ $(i=4)$ gives $6.0$ and $1 \ M \ Na_2SO_4$ $(i=3)$ gives $3.0$. Wait,let us re-evaluate.
Actually,for $1 \ M \ Na_2SO_4$,$i=3$,$iC = 3$. For $1.5 \ M \ AlCl_3$,$i=4$,$iC = 6$.
Let us check the options again. $1 \ M \ NaCl$ $(i=2, iC=2)$ and $2 \ M$ Urea $(i=1, iC=2)$.
Since $i_1 C_1 = i_2 C_2 = 2$,the pair in option $A$ is isotonic.

(B) The osmotic pressure $(\pi)$ of a solution is given by the van't Hoff equation: $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For a very dilute solution,the molar concentration $C$ is approximately equal to the molality $m$ because the density of the solution is approximately equal to the density of the solvent.
However,looking at the relationship $\pi = iCRT$,$\pi$ is directly proportional to $C$ (molarity) and $T$ (temperature).
If we consider the definition of molarity $C = n/V$,where $n$ is the number of moles $(n = \text{mass}/\text{molecular mass})$,then $\pi = (i \times \text{mass} \times R \times T) / (\text{molecular mass} \times V)$.
Thus,for a fixed mass of solute and fixed volume,$\pi \propto 1/\text{molecular mass}$.