Solutions – NEET UG ChemistryPractice Questions & PYQs

The relationship incorporating mass and molar mass for freezing point depression is given by $\Delta T_f = i K_f \frac{w_B \times 1000}{m_B w_A}$. Using the molar mass of $\mathrm{Na}_2\mathrm{SO}_4$ as 142 g/mol along with the provided values, the van't Hoff factor is isolated as $i = \frac{\Delta T_f m_B w_A}{K_f w_B \times 1000}$. Plugging in these specific figures yields $\frac{3.82 \times 142 \times 45.0}{1.86 \times 5.00 \times 1000}$, which computes to 2.63.

Mole fraction of solute is given by $\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$. A 1.00 molal solution contains 1 mole of solute dissolved in 1000 g of water, and since the molar mass of water is 18 g/mol, the number of moles of water is 55.56. Substituting these values into the mole fraction expression yields $\frac{1}{1 + 55.56}$, which evaluates to 0.0177.

The relationship between osmotic pressure, volume, mass, and molar mass is established by the equation $\pi V = \frac{w}{M} R T$, which rearranges to solve for the molar mass as $M = \frac{w R T}{\pi V}$. Substituting the given values for mass, gas constant, temperature, pressure, and volume into this formula produces the calculation $M = \frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-1} \times 0.2}$, which evaluates to a molar mass of 61038 g mol^-1.

Freezing point depression depends on particle concentration via the formula $\Delta T_f = i K_f m$, where the van't Hoff factor for a partially ionized weak acid is determined by $i = 1 + \alpha$. Given that the degree of ionization is 30 percent, the van't Hoff factor calculates to $1 + 0.3 = 1.3$, leading to the depression expression $\Delta T_f = 1.3 \times 1.86 \times 0.1$. Evaluating this product gives 0.24 degrees Celsius, which when subtracted from the standard freezing point results in -0.24 degrees Celsius.

Vapor pressure of a solution increases when the concentration of solute particles decreases because solvent molecules can escape more easily into the vapor phase. Adding water to the aqueous KI solution dilutes it, thereby lowering the overall solute concentration and raising the vapor pressure back toward that of pure solvent. In contrast, adding ionic compounds increases the particle count and further lowers the vapor pressure through colligative depression.

Freezing point depression is calculated using the relation $\Delta T_f = K_f m$, where molality represents the moles of solute per kilogram of solvent. The molality is determined by dividing the moles of sucrose, obtained from its given mass and molar mass, by the mass of water in kilograms, and multiplying this value by the cryoscopic constant. Subtracting the resulting temperature depression from the normal freezing point of pure water establishes the final freezing point of the solution as -0.520 °C.

The depression in freezing point is related to the van't Hoff factor through the equation $\Delta T_f = i K_f m$. Substituting the given freezing point depression, molality, and cryoscopic constant into the formula gives $0.00732 = i \times 1.86 \times 0.0020$. Solving this expression results in a van't Hoff factor of approximately 2, indicating that each mole of the ionic compound dissociates to produce 2 moles of ions in water.

The extent of dissociation of a weak electrolyte directly determines its van't Hoff factor through the relation $i = 1 + \alpha$. Substituting the twenty percent ionization degree gives $i = 1 + 0.2 = 1.2$. Applying this factor to the freezing point depression equation $\Delta T_f = i K_f m$ with a molality of 0.5 and cryoscopic constant 1.86 yields a depression of $1.2 \times 1.86 \times 0.5 = 1.116$ K, which rounds to 1.12 K.

Solutions having identical osmotic pressures at the same temperature are isotonic, meaning their molar concentrations are equal according to the relation $\frac{w_1}{M_1 V_1} = \frac{w_2}{M_2 V_2}$. Urea has a concentration of 10 g per liter and a molar mass of 60 g per mole, while the non-volatile solute has a concentration of 50 g per liter derived from its 5 percent mass-to-volume ratio. Equating the molar concentrations of both solutions gives $\frac{10}{60} = \frac{50}{M}$, which simplifies to yield a molar mass of 300 g per mole for the non-volatile solute.

Osmosis is a dynamic equilibrium process where solvent molecules pass through a semipermeable membrane in both directions simultaneously. Because real membranes allow bidirectional solvent exchange driven by chemical potential differences, water molecules cross the membrane from both sides, but at unequal rates resulting in net osmotic transfer.

Colligative property calculations for freezing point depression rely on the relationship $\Delta T_f = \frac{K_f \cdot w \cdot 1000}{M \cdot W}$, where $w$ is solute mass, $W$ is solvent mass, and $M$ is molar mass. Substituting the given values of 1.00 g for solute mass, 250 for molar mass, 51.2 g for benzene mass, and 5.12 for the cryoscopic constant into the expression gives $\Delta T_f = \frac{5.12 \times 1.00 \times 1000}{250 \times 51.2}$. Simplifying this calculation results in a freezing point depression of 0.4 K.

In non-ideal solutions, weaker intermolecular forces between unlike molecules compared to like molecules disrupt hydrogen bonding and lead to a higher escaping tendency of both components. Acetone and ethanol mixtures experience such weakened dipole-dipole interactions because acetone molecules interrupt the ethanol hydrogen-bonded network. Consequently, the total vapor pressure of the mixture exceeds the value predicted by Raoult's law, manifesting as a positive deviation.

Solution concentration and dilution are related through the molarity formula and the conservation principle $M_1 V_1 = M_2 V_2$. Calculating the molarity of the concentrated acid first requires finding that 100 g of solution has a volume of 55.55 mL, leading to a stock concentration of 18.02 M. Substituting the stock molarity, target molarity, and target volume into the dilution equation gives $V_1 = \frac{0.1 \times 1000}{18.02}$, which evaluates to a required volume of 5.55 mL.

The composition of a vapor phase above a liquid mixture is determined by combining Raoult's law and Dalton's law, where the vapor mole fraction is the ratio of the component's partial pressure to the total vapor pressure. Given a $1:4$ mole ratio, the liquid mole fractions are $\frac{1}{5}$ for pentane and $\frac{4}{5}$ for hexane. The partial pressure of pentane is calculated as $440 \times \frac{1}{5}$, which equals $88$ mm Hg, while the total pressure is the sum of both partial pressures amounting to $184$ mm Hg. Dividing the partial pressure of pentane by the total vapor pressure yields $\frac{88}{184}$, resulting in $0.478$.

A one molal solution contains $1$ mole of solute dissolved in $1000 \text{ g}$ of water. The number of moles of water is determined by dividing this mass by its molar mass, yielding $\frac{1000}{18} \text{ moles}$, which equals $55.56 \text{ moles}$. The mole fraction of the solute is then calculated using the formula $X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$. Substituting the values gives $\frac{1}{1 + 55.56}$, resulting in $0.018$.

The total vapor pressure of an ideal binary solution is determined by Raoult's law, where the total pressure is the sum of the partial pressures of each volatile component, represented by $P_{\text{total}} = p^\circ_{P} x_{P} + p^\circ_{Q} x_{Q}$. The individual mole fractions are found by dividing each component's moles by the total moles, giving $\frac{3}{5}$ for component $P$ and $\frac{2}{5}$ for component $Q$. Substituting these values and the pure vapor pressures of $80$ torr and $60$ torr into the equation yields $80 \times \frac{3}{5} + 60 \times \frac{2}{5}$, which simplifies to $48 + 24$, totaling $72$ torr.

Colligative properties like boiling point elevation and freezing point depression are both directly proportional to the molality of the solution, meaning their ratio depends solely on the respective cryoscopic and ebullioscopic constants expressed as $\frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b}$. Given a boiling point elevation of $0.18^\circ\text{C}$ along with $K_f = 1.86$ and $K_b = 0.512$, the freezing point depression can be found by substituting these values into the proportionality expression. Solving for $\Delta T_f$ gives $\frac{1.86}{0.512} \times 0.18$, which equals $0.654^\circ\text{C}$. Since freezing point is depressed below the normal freezing point of water, the resulting temperature is $-0.654^\circ\text{C}$.

The relation between osmotic pressure, concentration, and molecular mass is established through the gas equation adaptation $\pi = \frac{n}{V} R T$, where the number of moles represents mass divided by molar mass. Substituting $n = \frac{m_2}{M_2}$ into this expression and isolating the molar mass term rearranges the variables into the product of mass, gas constant, and temperature divided by the product of pressure and volume. Expressing this algebraic rearrangement in terms of mass per unit volume yields the required formulation.

The total moles of solute are calculated by combining the moles from both individual solutions, resulting in $2.5 \times 1$ plus $3 \times 0.5$, which equals $4$ moles. Adding the individual volumes together gives a combined volume of $5.5$ litres for the resulting mixture. Dividing these total moles by the total volume yields a resultant molarity of $0.73$ molar.

Raoult's law is perfectly obeyed by ideal solutions where the thermodynamic environment of the molecules remains completely unaltered upon mixing. For this ideal behavior to occur, the intermolecular attractive forces between unlike molecules must be identical in magnitude to the forces between like molecules. Consequently, no enthalpy change or volume change takes place when the components are combined.