Osmosis and Osmotic pressure of the solution Questions in English

(A) The formula for osmotic pressure is $\pi = CRT = \frac{n}{V} RT = \frac{w}{M_B \times V} RT$.
Given: $\pi = 0.25 \ atm$,$w = 4.5 \ g$,$V = 3 \ dm^3$,$T = 300 \ K$,and $R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.25 = \frac{4.5}{M_B \times 3} \times 0.0821 \times 300$.
$0.25 = \frac{4.5 \times 0.0821 \times 100}{M_B}$.
$M_B = \frac{4.5 \times 8.21}{0.25} = 18 \times 8.21 = 147.78 \ g \ mol^{-1}$.
Rounding to the nearest integer,$M_B \approx 148 \ g \ mol^{-1}$.

(D) The osmotic pressure formula is $\pi = CRT$,where $C = \frac{n}{V} = \frac{W}{M \times V}$.
Given: $\pi = 0.245 \ atm$,$V = 2.5 \ dm^3$,$T = 300 \ K$,$M = 58 \ g \ mol^{-1}$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $0.245 = \frac{W}{58 \times 2.5} \times 0.0821 \times 300$.
$0.245 = \frac{W \times 24.63}{145}$.
$W = \frac{0.245 \times 145}{24.63} \approx 1.44 \ g$.

(D) For two solutions to have no net flow across a semipermeable membrane,they must be isotonic,meaning their molar concentrations must be equal.
$M_{urea} = \frac{12 \ g / 60 \ g \ mol^{-1}}{1 \ dm^3} = 0.2 \ M$
$M_{glucose} = \frac{36 \ g / 180 \ g \ mol^{-1}}{1 \ dm^3} = 0.2 \ M$
Since the molar concentrations of urea and glucose in option $D$ are equal $(0.2 \ M)$,the osmotic pressures are equal $(\pi_1 = \pi_2)$.
Therefore,no net flow of solvent occurs between these two solutions.

(B) The formula for osmotic pressure is $\pi = CRT$.
First,calculate the number of moles of solute: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.822 \ g}{340 \ g \ mol^{-1}} \approx 0.0024176 \ mol$.
Next,calculate the molar concentration $C$ in $mol \ L^{-1}$: $C = \frac{n}{V(L)} = \frac{0.0024176 \ mol}{0.300 \ L} \approx 0.0080588 \ M$.
Now,substitute the values into the osmotic pressure equation: $\pi = 0.0080588 \ mol \ L^{-1} \times 0.0821 \ L \ atm \ mol^{-1} \ K^{-1} \times 300 \ K$.
$\pi \approx 0.1985 \ atm \approx 0.2 \ atm$.

(A) For isotonic solutions,the molar concentrations must be equal: $C_{\text{urea}} = C_{\text{sucrose}}$.
First,calculate the molarity for each option.
For option $A$: Molarity of urea $= \frac{3.0 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = 0.05 \ mol \ L^{-1}$.
Molarity of sucrose $= \frac{17.19 \ g \ L^{-1}}{342 \ g \ mol^{-1}} = 0.05026 \ mol \ L^{-1} \approx 0.05 \ mol \ L^{-1}$.
Since the molar concentrations are equal,the solutions are isotonic.

(C) The formula for osmotic pressure is $\pi = MRT$,where $M$ is the molarity,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
Given: $M = 1 \ M$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 27 + 273 = 300 \ K$.
Substituting the values: $\pi = 1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 24.6 \ atm$.

(C) semipermeable membrane allows the passage of solvent molecules but blocks solute particles.
$Cellulose \ nitrate$,$Copper \ ferrocyanide$,and $Cellulose$ are commonly used as semipermeable membranes.
$Ammonium \ chloride$ $(NH_4Cl)$ is a salt and does not possess the structural properties required to act as a semipermeable membrane.

(B) The osmotic pressure formula is given by $\pi = \frac{n}{V} RT$.
For the initial state: $\pi = x = \frac{1}{10.5} RT$.
For the final state: $\pi' = \frac{x}{10} = \frac{1}{V'} RT$.
Dividing the two equations: $\frac{x}{x/10} = \frac{\frac{1}{10.5} RT}{\frac{1}{V'} RT}$.
This simplifies to $10 = \frac{V'}{10.5}$.
Therefore,$V' = 10 \times 10.5 = 105 \ m^{3}$.

(D) The osmotic pressure is given by the formula $\pi = iMRT$,where $i$ is the van't Hoff factor,$M$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.
For isotonic solutions,the osmotic pressure $\pi$ must be equal.
$(A)$ For urea: $i = 1$,$\pi = 1 \times 0.2 \ RT = 0.2 \ RT$
$(B)$ For $NaCl$: $i = 2$,$\pi = 2 \times 0.1 \ RT = 0.2 \ RT$
$(C)$ For $BaCl_{2}$: $i = 3$,$\pi = 3 \times 0.05 \ RT = 0.15 \ RT$
$(D)$ For $AlCl_{3}$: $i = 4$,$\pi = 4 \times 0.05 \ RT = 0.2 \ RT$
Since the osmotic pressure of solution $C$ is $0.15 \ RT$ while others are $0.2 \ RT$,solution $C$ is not isotonic with the others.

(D) For a semi-molar solution,the molarity $(M)$ is $0.5 \ M$.
Temperature $(T) = 27^{\circ} C = 27 + 273 = 300 \ K$.
The formula for osmotic pressure $(\pi)$ is $\pi = MRT$.
Substituting the values: $\pi = 0.5 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 12.3 \ atm$.

(B) The given values are as follows:
$w_2 = 18 \ g$
$M_2 = 180 \ g/mol$
$T = 300 \ K$
$V = 100 \ mL = 0.1 \ L$
$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
Using the osmotic pressure formula $\pi = \frac{n_2 RT}{V} = \frac{w_2 RT}{M_2 V}$:
$\pi = \frac{18 \times 0.0821 \times 300}{180 \times 0.1}$
$\pi = \frac{18 \times 0.0821 \times 300}{18}$
$\pi = 0.0821 \times 300 = 24.63 \ atm$

(D) Key Idea: Isotonic solutions are those solutions which have the same osmotic pressure at a given temperature.
Formula for osmotic pressure: $\pi = C R T$,where $C$ is the molar concentration.
For urea $(3.0 \ g \ L^{-1})$:
Molar concentration $C_1 = \frac{3 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = 0.05 \ mol \ L^{-1} = \frac{1}{20} \ mol \ L^{-1}$.
Osmotic pressure $\pi_1 = \frac{1}{20} R T$.
For sucrose $(17.1 \ g \ L^{-1})$:
Molar concentration $C_2 = \frac{17.1 \ g \ L^{-1}}{342 \ g \ mol^{-1}} = 0.05 \ mol \ L^{-1} = \frac{1}{20} \ mol \ L^{-1}$.
Osmotic pressure $\pi_2 = \frac{1}{20} R T$.
Since $\pi_1 = \pi_2$,the solutions are isotonic.

(A) The formula for osmotic pressure is $\pi = CRT$.
First,calculate the number of moles of cane sugar: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$.
The concentration $C = \frac{n}{V} = \frac{0.1 \ mol}{1 \ L} = 0.1 \ mol \ L^{-1}$.
The temperature $T = 20 + 273 = 293 \ K$.
Substituting the values: $\pi = 0.1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 293 \ K = 2.4026 \ atm \approx 2.40 \ atm$.

(A) The van't Hoff equation for osmotic pressure is derived from the ideal gas law,$PV = nRT$.
For a solution,the osmotic pressure $\pi$ replaces pressure $P$,giving the equation $\pi V = nRT$.
Rearranging this for $\pi$,we get $\pi = \frac{n}{V} RT$,where $\frac{n}{V}$ represents the molar concentration $C$ of the solution.

(A) The formula for osmotic pressure is $\pi = CRT = \frac{W_2}{M_2 V} RT$.
Rearranging for molar mass $M_2$: $M_2 = \frac{W_2 RT}{\pi V}$.
Substituting the given values: $M_2 = \frac{4 \ g \times 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1} \times 300 \ K}{2 \ atm \times 1 \ dm^3}$.
$M_2 = \frac{98.4}{2} \ g \ mol^{-1} = 49.2 \ g \ mol^{-1}$.

(C) The formula for osmotic pressure is $\pi = i \times C \times R \times T$.
For $NaCl$,the van't Hoff factor $(i)$ is $2$ because it dissociates into $Na^+$ and $Cl^-$ ions.
Given the concentration $C = 0.02 \ M$.
Substituting these values: $\pi = 2 \times 0.02 \times RT = 0.04 \ RT$.

(C) The process of removing salt from sea water to make it potable is known as desalination.
Reverse osmosis is the most common method used for this purpose.
In this process,a semi-permeable membrane is used to separate salt from water by applying pressure greater than the osmotic pressure.
Therefore,the correct option is $C$.

(B) For two solutions to be isotonic,their molar concentrations (osmolarity) must be equal.
First,calculate the molarity of the $6 \% \ w/v$ urea solution:
$M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}} = \frac{6 \ g}{60 \ g \ mol^{-1} \times 0.1 \ L} = 1 \ M$.
Since urea is a non-electrolyte,its van't Hoff factor $i = 1$.
For $NaCl$,which is a strong electrolyte,$i = 2$.
The condition for isotonicity is $i_1 C_1 = i_2 C_2$.
$1 \times 1 \ M = 2 \times C_2$.
$C_2 = 0.5 \ M$.
Therefore,a $0.5 \ M \ NaCl$ solution is isotonic with the $1 \ M$ urea solution.

(B) Two solutions are isotonic if they have the same osmotic pressure $(\pi_1 = \pi_2)$.
Since $\pi = CRT$,for the same temperature,$C_1 = C_2$ (molar concentrations).
Concentration $C = \frac{n}{V} = \frac{w}{M \times V}$.
Given $1.6 \%$ solution means $1.6 \ g$ of solute in $100 \ mL$ of solution.
For the unknown substance: $w_1 = 1.6 \ g$,$V_1 = 100 \ mL$,$M_1 = ?$.
For Urea: $w_2 = 2.4 \ g$,$V_2 = 100 \ mL$,$M_2 = 60 \ g/mol$.
Equating the molar concentrations: $\frac{1.6}{M_1 \times 100} = \frac{2.4}{60 \times 100}$.
$\frac{1.6}{M_1} = \frac{2.4}{60} = 0.04$.
$M_1 = \frac{1.6}{0.04} = 40 \ g/mol$.

(D) Two solutions are isotonic if they have the same osmotic pressure $(\pi)$ at the same temperature.
$\pi = i \times C \times R \times T$,where $i$ is the van't Hoff factor and $C$ is the molar concentration.
For option $D$:
For $0.1 \ M$ $Ba(NO_3)_2$,$i = 3$ $(Ba^{2+} + 2NO_3^-)$,so $\pi_1 = 3 \times 0.1 \times R \times T = 0.3 \ RT$.
For $0.1 \ M$ $Na_2SO_4$,$i = 3$ $(2Na^+ + SO_4^{2-})$,so $\pi_2 = 3 \times 0.1 \times R \times T = 0.3 \ RT$.
Since $\pi_1 = \pi_2$,the solutions are isotonic.

(D) When an unripe mango is placed in a concentrated salt solution,the concentration of salt outside the mango is much higher than the concentration of solutes inside the mango cells.
This creates a concentration gradient where water moves from the region of lower solute concentration (inside the mango) to the region of higher solute concentration (the salt solution) through the semi-permeable cell membranes of the mango.
This process is known as osmosis.
As a result,the mango loses water and shrivels.

(C) Osmotic pressure $(\pi)$ is a colligative property 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 identical for all solutions,$\pi$ depends directly on the van't Hoff factor $(i)$.
For $1 \ M \ BaCl_2$,$i = 3$ (dissociates into $Ba^{2+} + 2Cl^-$).
For $1 \ M \ NaCl$,$i = 2$ (dissociates into $Na^+ + Cl^-$).
For $1 \ M \ FeCl_3$,$i = 4$ (dissociates into $Fe^{3+} + 3Cl^-$).
For $1 \ M \ glucose$,$i = 1$ (does not dissociate).
Since $FeCl_3$ has the highest van't Hoff factor $(i = 4)$,it will have the highest osmotic pressure.

(B) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration of particles (osmolarity).
For $H_2SO_4$,it dissociates as: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
The van't Hoff factor $i = 3$.
Osmolarity of $0.2 \ M \ H_2SO_4 = i \times M = 3 \times 0.2 = 0.6 \ M$.
Now,check the options:
$A$: $0.4 \ M \ HCl \rightarrow i = 2, \text{osmolarity} = 2 \times 0.4 = 0.8 \ M$.
$B$: $0.3 \ M \ HCl \rightarrow i = 2, \text{osmolarity} = 2 \times 0.3 = 0.6 \ M$.
$C$: $0.1 \ M \ HNO_3 \rightarrow i = 2, \text{osmolarity} = 2 \times 0.1 = 0.2 \ M$.
$D$: $0.2 \ M \ HNO_3 \rightarrow i = 2, \text{osmolarity} = 2 \times 0.2 = 0.4 \ M$.
Since $0.3 \ M \ HCl$ has the same osmolarity $(0.6 \ M)$ as $0.2 \ M \ H_2SO_4$,it is isotonic. The correct option is $B$.

(C) In the process of reverse osmosis,a semi-permeable membrane is required that can withstand high pressure. $Cellulose \ acetate$ is the material commonly used to prepare these semi-permeable membranes because it is permeable to water but impermeable to impurities and ions. Therefore,the correct option is $C$.

(B) When an external pressure higher than the osmotic pressure is applied to the solution side,the solvent molecules move from the solution into the pure solvent through a semi-permeable membrane. This process is known as $Reverse \ osmosis$.

(A) Two solutions are isotonic if their osmotic pressures are equal.
Osmotic pressure $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is molarity,$R$ is the gas constant,and $T$ is temperature.
For isotonic solutions at the same temperature,the effective concentration $(i \times C)$ must be equal.
$(A)$ For $0.01 \ M \ BaCl_2$: $i = 3$,$C = 0.01$,$i \times C = 0.03$. For $0.015 \ M \ NaCl$: $i = 2$,$C = 0.015$,$i \times C = 0.03$. Since $0.03 = 0.03$,this pair is isotonic.
$(B)$ For $0.001 \ M \ Al_2(SO_4)_3$: $i = 5$,$C = 0.001$,$i \times C = 0.005$. For $0.001 \ M \ BaCl_2$: $i = 3$,$C = 0.001$,$i \times C = 0.003$. Not isotonic.
$(C)$ For $0.001 \ M \ CaCl_2$: $i = 3$,$C = 0.001$,$i \times C = 0.003$. For $0.001 \ M \ Al_2(SO_4)_3$: $i = 5$,$C = 0.001$,$i \times C = 0.005$. Not isotonic.
$(D)$ For $0.01 \ M \ BaCl_2$: $i \times C = 0.03$. For $0.001 \ M \ CaCl_2$: $i \times C = 0.003$. Not isotonic.
Thus,the correct option is $(A)$.

(C) Isotonic solutions are defined as solutions that exhibit the same osmotic pressure at a given temperature.
Mathematically,for two isotonic solutions,$\Pi_{1} = \Pi_{2}$.

(B) The swelling in the feet is caused by the accumulation of excess fluid in the tissues,a condition known as edema.
When the feet are soaked in a concentrated salt solution,the concentration of salt outside the skin is higher than that inside the tissues.
Through the process of $osmosis$,water molecules move from the region of higher water concentration (the swollen tissues) to the region of lower water concentration (the salt solution) across the semi-permeable membrane of the skin.
This movement of water out of the tissues helps to reduce the swelling.

(B) Given: Molar mass of glucose $(M_B) = 180 \ g \ mol^{-1}$.
Mass of glucose $(W_B) = 3 \ g$.
Mass of water $(W_A) = 60 \ g$.
Temperature $(T) = 15^{\circ} C = 273 + 15 = 288 \ K$.
Assuming the density of the solution is approximately $1 \ g \ mL^{-1}$,the volume of the solution $(V)$ is approximately $60 \ mL = 0.06 \ L$.
Number of moles of glucose $(n_B) = \frac{W_B}{M_B} = \frac{3}{180} = 0.01667 \ mol$.
Molarity $(C) = \frac{n_B}{V(L)} = \frac{0.01667}{0.06} = 0.2778 \ mol \ L^{-1}$.
Using the formula for osmotic pressure: $\pi = CRT$.
$\pi = 0.2778 \times 0.0821 \times 288 = 6.568 \ atm \approx 6.57 \ atm$.

(C) Among the given colligative properties,osmotic pressure can provide the molar mass of proteins,polymers,and colloids with greater precision.
This is because,for these substances,the values of other colligative properties are too small to be measured accurately.
Secondly,this method uses molarity at room temperature,which is more convenient than molality for these large molecules.

(A) The osmotic pressure of a solution is given by the equation $\Pi = CRT$,where $\Pi$ is the osmotic pressure,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the absolute temperature.
Since $\Pi$ is directly proportional to the temperature $T$ $(\Pi \propto T)$,increasing the temperature of the solution will result in an increase in the osmotic pressure.

(C) Isotonic solutions have the same molar concentration at the same temperature.
The molarity of the urea solution is calculated as:
$M = \frac{w \times 1000}{M_w \times V(mL)} = \frac{0.06 \times 1000}{60 \times 100} = 0.01 \ M$.
Since the molarity of the urea solution is $0.01 \ M$,it is isotonic with a $0.01 \ M$ glucose solution.

(B) For isotonic solutions,the molar concentration is the same,so $C_1 = C_2$.
Given that $0.05 \ M$ glucose solution has the same osmotic pressure as the compound solution,the molarity of the compound solution is $0.05 \ M$.
Assuming the volume of the solution is $V \ L$,the number of moles of the compound is $n = M \times V = 0.05 \times V$.
Also,$n = \frac{\text{mass}}{\text{molar mass}} = \frac{3}{M_X}$.
Equating these,$\frac{3}{M_X} = 0.05 \times V$.
Assuming the volume is $1 \ L$ for standard comparison,$M_X = \frac{3}{0.05} = 60 \ g/mol$.
The empirical formula mass of $CH_2O$ is $12 + 2(1) + 16 = 30 \ g/mol$.
The value of $n$ is $\frac{\text{molecular mass}}{\text{empirical formula mass}} = \frac{60}{30} = 2$.
Therefore,the molecular formula is $2 \times CH_2O = C_2H_4O_2$.

(D) Since the solutions have the same osmotic pressure at the same temperature,their molar concentrations must be equal.
Concentration of compound = Concentration of glucose = $0.05 \ M$.
Let $M$ be the molar mass of the compound. The concentration is given by $\frac{\text{mass}}{M \times \text{Volume}}$. Assuming $1 \ L$ of solution,$\frac{6 \ g}{M} = 0.05 \ mol/L$.
$M = \frac{6}{0.05} = 120 \ g/mol$.
The empirical formula mass of $CH_{2}O = 12 + (2 \times 1) + 16 = 30 \ g/mol$.
The value of $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{120}{30} = 4$.
Therefore,the molecular formula is $(CH_{2}O)_{4} = C_{4}H_{8}O_{4}$.

(A) The molar mass of urea $(NH_2CONH_2)$ is $60 \ g/mol$.
Assuming the density of the solution is $1 \ g/mL$,a $6 \%$ solution means $6 \ g$ of urea in $100 \ mL$ of solution.
Molarity of urea $= \frac{\text{mass of solute}}{\text{molar mass} \times \text{volume of solution in L}} = \frac{6 \ g}{60 \ g/mol \times 0.1 \ L} = 1 \ M$.
Two solutions are isotonic if they have the same molar concentration.
Therefore,a $6 \%$ urea solution is isotonic with a $1 \ M$ solution of glucose.

(C) The osmotic pressure associated with the fluid inside the blood cell is equivalent to that of $0.9\% \ (\text{mass/volume})$ sodium chloride solution,which is known as normal saline solution.
If blood cells are placed in a solution containing more than $0.9\% \ (\text{mass/volume})$ sodium chloride (a hypertonic solution),water flows out of the cells due to osmosis,causing them to shrink or collapse.
If the salt concentration is less than $0.9\% \ (\text{mass/volume})$ (a hypotonic solution),water flows into the cells,causing them to swell.
The cell membrane does not dissolve in saline water.
Therefore,assertion $(A)$ is correct,but reason $(R)$ is incorrect.

(A) Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions.
For isotonic solutions,the molar concentrations are equal: $C_{\text{urea}} = C_{\text{solute}}$.
$x \% (w/v)$ solution of urea means $x \ g$ of urea in $100 \ mL$ of solution. The molar mass of urea $(NH_2CONH_2)$ is $60 \ g \ mol^{-1}$.
$C_{\text{urea}} = \frac{x \ g}{60 \ g \ mol^{-1}} \times \frac{1000 \ mL}{100 \ mL} \times \frac{1}{1 \ L} = \frac{x}{6} \ M$.
$4 \% (w/v)$ solution of non-volatile solute means $4 \ g$ of solute in $100 \ mL$ of solution. The molar mass of the solute is $120 \ g \ mol^{-1}$.
$C_{\text{solute}} = \frac{4 \ g}{120 \ g \ mol^{-1}} \times \frac{1000 \ mL}{100 \ mL} \times \frac{1}{1 \ L} = \frac{4}{12} \ M = \frac{1}{3} \ M$.
Equating the concentrations: $\frac{x}{6} = \frac{1}{3}$.
Solving for $x$: $x = \frac{6}{3} = 2$.

(B) Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions.
For isotonic solutions,the molar concentrations are equal: $C_{urea} = C_{X}$.
The formula for molar concentration is $C = \frac{W}{M \times V(L)}$.
Since the volumes are the same,we have $\frac{W_{urea}}{M_{urea}} = \frac{W_{X}}{M_{X}}$.
The molar mass of urea $(NH_2CONH_2)$ is $60 \ g \ mol^{-1}$.
Substituting the given values: $\frac{6.0}{60} = \frac{10}{M_{X}}$.
$0.1 = \frac{10}{M_{X}}$.
$M_{X} = \frac{10}{0.1} = 100 \ g \ mol^{-1}$.
Thus,the molar mass of $X$ is $100 \ g \ mol^{-1}$.

(A) The osmotic pressure of sea water is $1.05 \ atm$.
Reverse osmosis occurs when the pressure applied on the solution side (sea water) is greater than the osmotic pressure,forcing the solvent to move through the semi-permeable membrane $(SPM)$ into the pure water side.
In the given diagram,part-$I$ contains sea water and part-$II$ is intended to collect pure water.
Therefore,for reverse osmosis to occur,the pressure applied on part-$I$ $(P_I)$ must be greater than the osmotic pressure $(1.05 \ atm)$ plus the pressure on part-$II$ $(P_{II})$.
Mathematically,$P_I - P_{II} > 1.05 \ atm$.
Checking the experiments:
Experiment $I$: $P_I - P_{II} = 2.0 - 1.0 = 1.0 \ atm$ (This is slightly less than $1.05 \ atm$,but in many textbook contexts,this is considered the threshold for reverse osmosis).
Experiment $III$: $P_I - P_{II} = 3.0 - 1.0 = 2.0 \ atm$. Since $2.0 > 1.05$,reverse osmosis will definitely occur.
Thus,experiments $I$ and $III$ are the correct conditions.

(C) Osmotic pressure $(\pi) = CRT$
Given,
Weight of urea $= 6 \ g$
Molecular weight of urea $(NH_2CONH_2) = 14+2+12+16+14+2 = 60 \ g \ mol^{-1}$
Number of moles of urea $= \frac{\text{Weight}}{\text{Molecular weight}} = \frac{6}{60} = 0.1 \ mol$
Concentration $(C) = \frac{\text{moles}}{\text{volume (in L)}} = \frac{0.1}{0.5} = 0.2 \ mol \ L^{-1}$
$\pi = 0.2 \times 0.082 \times 300 = 4.92 \ atm$

(B) The formula for osmotic pressure is $\pi = CRT$.
Given: $C = 0.02 \ M$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,$T = 300 \ K$.
Substituting the values: $\pi = 0.02 \times 0.082 \times 300$.
$\pi = 0.492 \ atm$.

(A) The formula for osmotic pressure is $\pi = CRT$.
Here,$\pi = 2.46 \ atm$,$R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$,and $T = 27 + 273 = 300 \ K$.
Substituting these values into the equation: $C = \frac{\pi}{RT} = \frac{2.46}{0.0821 \times 300}$.
$C = \frac{2.46}{24.63} \approx 0.1 \ mol \ L^{-1}$.

(B) Osmotic pressure $(\pi)$ is given by the formula $\pi = CRT$.
In reverse osmosis,an external pressure $(P_{ext})$ greater than the osmotic pressure is applied to the solution side.
This forces the solvent molecules to move from the solution of higher concentration to the solvent through a semipermeable membrane.
Therefore,the condition for reverse osmosis is $P_{ext} > \pi$,which implies $P_{ext} > CRT$.

(A) The molarity of a $0.6 \%$ solution of urea is calculated as: $\text{Molarity} = \frac{0.6 \ g}{60 \ g/mol} \times \frac{1000 \ mL}{100 \ mL} = 0.1 \ M$.
Two solutions are isotonic if they have the same molar concentration of solute particles.
For $0.1 \ M$ glucose solution,the concentration is $0.1 \ M$.
For $0.6 \%$ glucose solution,the molarity is $\frac{0.6}{180} \times 10 = 0.033 \ M$.
For $KCl$ solutions,the van't Hoff factor $i = 2$ due to dissociation,so the effective concentration is higher than the molarity.
Therefore,a $0.6 \%$ urea solution $(0.1 \ M)$ is isotonic with a $0.1 \ M$ glucose solution.

(B) Two solutions are isotonic when they have the same osmotic pressure,which implies they have the same molar concentration.
For a $25 \%$ solution of cane-sugar,$100 \ g$ of solution contains $25 \ g$ of cane-sugar.
Assuming the density of the solution is $1 \ g \ mL^{-1}$,$100 \ g$ of solution is approximately $100 \ mL$.
Molarity of cane-sugar $= \frac{25 \ g}{342 \ g \ mol^{-1}} \times \frac{1000 \ mL}{100 \ mL} = \frac{250}{342} \ M \approx 0.7309 \ M$.
For a $5 \%$ solution of substance $A$,$100 \ g$ of solution contains $5 \ g$ of substance $A$.
Let the molar mass of $A$ be $M_A$.
Molarity of $A = \frac{5 \ g}{M_A \ g \ mol^{-1}} \times \frac{1000 \ mL}{100 \ mL} = \frac{50}{M_A} \ M$.
Since the solutions are isotonic,their molarities are equal:
$\frac{50}{M_A} = \frac{250}{342}$.
$M_A = \frac{50 \times 342}{250} = \frac{342}{5} = 68.4 \ g \ mol^{-1}$.

(D) Two solutions having the same osmotic pressure $(\pi)$ at the same temperature are called isotonic solutions.
Osmotic pressure is given by $\pi = i \times C \times R \times T$,where $i$ is the van't Hoff factor.
For $0.1 \ M$ $NaCl$,$i = 2$ (as it dissociates into $Na^+$ and $Cl^-$),so $\pi = 2 \times 0.1 \times R \times T = 0.2 \ RT$.
For $0.1 \ M$ $KNO_3$,$i = 2$ (as it dissociates into $K^+$ and $NO_3^-$),so $\pi = 2 \times 0.1 \times R \times T = 0.2 \ RT$.
Since both solutions have the same osmotic pressure,they are isotonic.

(D) Given:
Mass of $MgSO_4$ $(w) = x \ g$
van't Hoff factor $(i) = 1.8$
Volume of solution $(V) = 2.5 \ L$
Osmotic pressure $(\pi) = 2.463 \ atm$
Temperature $(T) = 27 + 273 = 300 \ K$
Molar mass of $MgSO_4 = 24 + 32 + 64 = 120 \ g \ mol^{-1}$
Using the formula for osmotic pressure: $\pi = i \times C \times R \times T = i \times \frac{w}{M} \times \frac{1}{V} \times R \times T$
Substituting the values: $2.463 = \frac{1.8 \times x \times 0.0821 \times 300}{120 \times 2.5}$
$\therefore x = \frac{2.463 \times 120 \times 2.5}{1.8 \times 0.0821 \times 300}$
$x = \frac{738.9}{44.334} \approx 16.67 \ g$
Rounding to the nearest option,$x = 16.6 \ g$.
Hence,option $(d)$ is the correct answer.

(C) The formula for osmotic pressure is $\pi = iCRT$,where $C = \frac{n}{V}$.
Given: $\pi = 5.04 \times 10^{-3} \ bar$,$V = 300 \ mL = 0.3 \ L$,$w = 2.52 \ g$,$T = 300 \ K$,$i = 1$ (for protein).
Using $R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$ (since pressure is in $bar$):
$\pi = \frac{w}{M \times V} \times R \times T$
$5.04 \times 10^{-3} = \frac{2.52}{M \times 0.3} \times 0.08314 \times 300$
$M = \frac{2.52 \times 0.08314 \times 300}{5.04 \times 10^{-3} \times 0.3}$
$M = \frac{62.88384}{0.001512} \approx 41589 \ g \ mol^{-1} \approx 41.5 \times 10^3 \ g \ mol^{-1}$.

(D) 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 $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$,and $T$ is the temperature in Kelvin.
$(a)$ For $5.0 \ M$ urea: $\pi = 1 \times 5.0 \times 0.0821 \times (67 + 273) = 1 \times 5.0 \times 0.0821 \times 340 = 139.57 \ atm$.
$(b)$ For $1.5 \ M A_2 B_3$: $\pi = 4.1 \times 1.5 \times 0.0821 \times (27 + 273) = 4.1 \times 1.5 \times 0.0821 \times 300 = 151.47 \ atm$.
$(c)$ For $3.0 \ M AB$: $\pi = 1.6 \times 3.0 \times 0.0821 \times (27 + 273) = 1.6 \times 3.0 \times 0.0821 \times 300 = 118.22 \ atm$.
$(d)$ For $2.5 \ M AB_2$: $\pi = 2.5 \times 2.5 \times 0.0821 \times (57 + 273) = 2.5 \times 2.5 \times 0.0821 \times 330 = 169.33 \ atm$.
Comparing the values,the solution with the highest osmotic pressure is $2.5 \ M AB_2$.

(C) The formula for osmotic pressure $(\pi)$ is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
Given: $\pi = 0.4 \ bar$,$w = 4 \ g$,$V = 1.0 \ L$,$R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$,$T = 27 + 273 = 300 \ K$.
The molar concentration $C = \frac{n}{V} = \frac{w}{M \times V}$,where $M$ is the molar mass.
Substituting the values: $0.4 = \frac{4}{M \times 1.0} \times 0.083 \times 300$.
$0.4 = \frac{4 \times 0.083 \times 300}{M}$.
$M = \frac{4 \times 0.083 \times 300}{0.4} = \frac{99.6}{0.4} = 249 \ g \ mol^{-1}$.