Solutions

(A) To calculate molality, we use the formula:

$ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} $

Moles of ethanoic acid $= \frac{2.5}{60}$ and mass of benzene $= 75 \text{ g} = 0.075 \text{ kg}$.

So, $ \text{Molality} = \frac{2.5/60}{0.075} = 0.556 \, \text{molal} $

Therefore, statement (A) is correct.

(B) To calculate molarity, we use the formula:

$ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution (in L)}} $

Moles of NaOH $= \frac{5}{40}$ and volume $= 450 \text{ mL} = 0.45 \text{ L}$.

So, $ \text{Molarity} = \frac{5/40}{0.45} = 0.278 \, \text{M} $

Hence, statement (B) is also correct.

(C) Aquatic animals find cold water more suitable because the solubility of gases increases when temperature decreases. From Henry’s law:

$ K_H \propto \text{Temperature} \quad \text{and} \quad K_H \propto \frac{1}{\text{Solubility}} $

This means higher temperature reduces gas solubility. So, statement (C) is correct.

(D) According to Henry’s law, the relationship between pressure and solubility is given by:

$ P = K_H X $

Here, $P \propto X$, i.e., solubility increases with an increase in pressure. So, the statement “solubility decreases with pressure” is incorrect.

(E) The mole fraction of component B in a binary mixture is given by:

$ x_B = \frac{n_B}{n_A + n_B} $

The given statement uses $n_A$ in the numerator, which is wrong. Hence, (E) is incorrect.

Acetone and chloroform show negative deviation from Raoult's law due to stronger H-bonding between acetone and chloroform mixture.

Hence, escaping tendency decreases

∴ Vapour pressure decreases

∴ Boiling point increases.

The boiling point elevation of an aqueous solution can be determined using the formula:

$ \Delta \mathrm{T}_{\mathrm{b}} = i \mathrm{~K}_{\mathrm{b}} \times \mathrm{m} $

Where:

$ \Delta \mathrm{T}_{\mathrm{b}} $ is the change in boiling point.

$ i $ is the van’t Hoff factor (number of particles the solute breaks into).

$ \mathrm{K}_{\mathrm{b}} $ is the ebullioscopic constant of the solvent.

$ \mathrm{m} $ is the molality of the solution.

For simplicity, we'll assume molarity equals molality. Here's the calculation for each solution:

0.01 M Na$_2$SO$_4$:

$ i = 3 $ (Na$_2$SO$_4$ dissociates into 2 Na$^+$ and 1 SO$_4^{2-}$)

$ i \times m = 3 \times 0.01 = 0.03 $

0.015 M C$_6$HundefinedO$_6$ (glucose):

$ i = 1 $ (glucose does not dissociate)

$ i \times m = 1 \times 0.015 = 0.015 $

0.01 M Urea:

$ i = 1 $ (urea does not dissociate)

$ i \times m = 1 \times 0.01 = 0.01 $

0.01 M KNO$_3$:

$ i = 2 $ (KNO$_3$ dissociates into 1 K$^+$ and 1 NO$_3^-$)

$ i \times m = 2 \times 0.01 = 0.02 $

The boiling point elevation is directly proportional to $ i \times m $. Thus, the solution with the highest $ i \times m $ value will have the highest boiling point.

Therefore, the 0.01 M Na$_2$SO$_4$ solution, with an $ i \times m $ value of 0.03, will have the highest boiling point.

$\bullet$ Humidity is a solution of liquid in gas

$\bullet$ Alloy is a solution of solid in solid

$\bullet$ Amalgam is a solution of liquid in solid

$\bullet$ Smoke is a solution of solid in gas

Given:

5 moles of liquid X

10 moles of liquid Y

The total moles in the solution are 15 (5 moles X + 10 moles Y).

The mole fraction of X ($X_x$) is:

$ \frac{5}{15} = \frac{1}{3} $

The mole fraction of Y ($X_y$) is:

$ \frac{10}{15} = \frac{2}{3} $

The vapor pressures of pure liquids are:

$P_x^{\circ} = 63$ torr for liquid X

$P_y^{\circ} = 78$ torr for liquid Y

Using Raoult's Law for an ideal solution, the total vapor pressure ($P_{\text{total}}$) is calculated as:

$ P_{\text{total}} = X_x \cdot P_x^{\circ} + X_y \cdot P_y^{\circ} $

Substitute the values:

$ P_{\text{total}} = \left(\frac{1}{3} \times 63\right) + \left(\frac{2}{3} \times 78\right) $

Calculate each term:

$ = 21 + 52 $

$ = 73 \text{ torr} $

The observed vapor pressure of the solution is 70 torr, which is lower than the calculated ideal pressure of 73 torr. This indicates a negative deviation from Raoult's Law because the observed vapor pressure is less than expected for an ideal solution. Therefore, the solution exhibits stronger intermolecular attractions than those in an ideal solution, leading to a lower vapor pressure.

For isotonic solutions [osmotic pressure must be equal]

$\begin{aligned} & \pi_1=\pi_2 \\ & C_1 R T=C_2 R T \\ & C_1=C_2 \\ & \frac{\mathrm{m}}{180 \times 1}=\frac{15}{60 \times 1} \quad \text { ( } \mathrm{m} \text { is the mass of glucose) } \\ & \mathrm{m}=\frac{180}{4}=45 \mathrm{~g} \end{aligned}$

According to Henry's law, the solubility of a gas in a liquid under constant temperature is directly proportional to the partial pressure of the gas above the liquid but is inversely proportional to the Henry's law constant $ (\mathrm{K}_\mathrm{H}) $ for the gas. Henry's law can be expressed as:

$\mathrm{c} = \frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}$

where $ \mathrm{c} $ is the concentration (or solubility) of the gas in the liquid, $ \mathrm{P} $ is the partial pressure of the gas, and $ \mathrm{K}_{\mathrm{H}} $ is Henry's law constant.

From the equation, it is clear that the solubility of the gas is inversely related to $ \mathrm{K}_{\mathrm{H}} $; if $ \mathrm{K}_{\mathrm{H}} $ increases, the solubility decreases, and vice versa.

In the given problem, you have the $ \mathrm{K}_{\mathrm{H}} $ values of the gases A, B, and C as follows:

• A: 145 kbar
• B: $ 2 \times 10^{-5} \mathrm{~kbar} $
• C: 35 kbar

Comparing the $ \mathrm{K}_{\mathrm{H}} $ values:

• Gas B has the lowest $ \mathrm{K}_{\mathrm{H}} $ and hence the highest solubility.
• Gas A, with the highest $ \mathrm{K}_{\mathrm{H}} $ among the three, will have the lowest solubility.
• Gas C has a $ \mathrm{K}_{\mathrm{H}} $ value less than A but greater than B, so its solubility will be lower than B but higher than A.

Therefore, the order of solubility of the gases in water from the highest to the lowest is:

B > C > A

Thus, the correct option is:

Option B: B > C > A

The relationship between the osmotic pressure $(\Pi)$ of a solution and its concentration $(c)$ can be derived from the van't Hoff equation for dilute solutions, which is given by:

$\Pi = cRT$

where:

• $\Pi$ is the osmotic pressure,
• $c$ is the concentration of the solution in moles per liter,
• $R$ is the ideal gas constant in appropriate units, and
• $T$ is the temperature in Kelvin.

According to the problem, the slope of the $\Pi$ vs $c$ plot is given as $25.73 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1}$ which corresponds to the product $RT$ from the van't Hoff equation. We are provided with the value of the gas constant $R = 0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$.

To find the temperature $T$, we use the following equation derived from the slope of the line:

$RT = 25.73 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1}$

To isolate $T$, we rearrange the equation:

$T = \frac{25.73 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1}}{0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}} \approx 310 \mathrm{~K}$

To convert this temperature from Kelvin to Celsius, we use the conversion formula:

$T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15$

$T_{\text{Celsius}} = 310 \mathrm{~K} - 273.15 \approx 36.85^{\circ} \mathrm{C}$

This value is closest to $37^{\circ} \mathrm{C}$, which corresponds to Option A. Therefore, the temperature at which the osmotic pressure measurement is done is approximately $37^{\circ} \mathrm{C}$.

$\mathrm{i\times M\downarrow \Rightarrow \Delta T_b \downarrow}$

According to Henry's Law,

p = K_Hx

Where 'p' is partial pressure of gas in vapour phase.

K_H is Henry's Law constant.

'x' is mole fraction of gas in liquid.

Higher the value of K_H at a given pressure, lower is the solubility of the gas in the liquid.

$\therefore$ Solubility : Ar < CO₂ < CH₄ < HCHO

The correct answer is Option D: from both sides of semipermeable membrane with unequal flow rates.

In osmosis, water (or any solvent) moves through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration. This semipermeable membrane allows the passage of solvent molecules but not solute molecules. The primary aim of this process is to equalize the solute concentrations on both sides of the membrane. However, it is essential to note that while water molecules predominantly move from the area of lower concentration to the area of higher concentration, there is still some movement of water from the higher concentration to the lower concentration side as well. This bidirectional movement occurs because water molecules are constantly in motion, but the net flow results in a higher volume of water moving to the side with higher solute concentration.

This process does not occur with equal flow rates because the driving force of osmosis is to reduce the concentration gradient between the two sides. Therefore, the flow rate is predominantly higher from the lower solute concentration side to the higher solute concentration side until an equilibrium is reached, at which point the net flow of water equalizes, but does not necessarily stop. This imbalance in flow rates demonstrates why Option D is the most accurate description of osmotic flow through a semipermeable membrane.