Q.Determine the amount of $CaCl_2$ ($i = 2.47$) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27$^\circ$C.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A water sample is contaminated with compound X (molar mass = $120\, \mathrm{g\,mol^{-1}}$). Its molality is $10^{-4}\ \mathrm{m}$. What is its concentration in ppm? (A) $120$ (B) $1200$ (C) $12$ (D) $12 \times 10^3$

   ›Reveal solutionSolution

   For dilute aqueous solutions, ppm by mass equals molality (in mol/kg) times molar mass (in g/mol). Here, $10^{-4} \times 120 = 0.012$ g solute per kg solvent, which is $12$ ppm.

   The idea is simple: ppm (parts per million) means grams of solute per million grams of solution. For a very dilute aqueous solution, the mass of the solution is almost exactly the mass of the water (since the solute mass is tiny). So we can treat "per kg of solvent" (molality) as "per kg of solution" without meaningful error.

   Molality $m = 10^{-4}\ \mathrm{mol\,kg^{-1}}$ means $10^{-4}$ moles of X are dissolved in 1 kg of water. Multiply by molar mass to get the mass of X in that 1 kg of water:

   $\text{mass of X} = 10^{-4}\ \mathrm{mol} \times 120\ \mathrm{g\,mol^{-1}} = 0.012\ \mathrm{g}$

   So 0.012 g of X is present in roughly 1 kg of solution. Now, ppm is defined as:

   $\text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6$

   Here, mass of solution ≈ 1000 g (since 1 kg water ≈ 1000 g, and solute mass is negligible). So:

   $$$$

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2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.2 moles of ammonia gas is present in a 5.0 L flask at 27 $^\circ$C. The pressure of it (in atm) using van der waals equation of state is ($a = 4.17\, \text{atm L}^2 \text{mol}^{-2}$, $b = 0.0371\, \text{L mol}^{-1}$ and $R = 0.0821\, \text{L atm K}^{-1} \text{mol}^{-1}$) (A) 9.32 (B) 4.66 (C) 13.98 (D) 2.33

   ›Reveal solutionSolution

   This problem asks us to calculate the pressure of ammonia gas using the van der Waals equation, which accounts for the non-ideal behavior of real gases. By substituting the given values for moles, volume, temperature, and van der Waals constants into the equation, we find the pressure to be approximately $\boxed{9.32 \text{ atm}}$.

   The ideal gas law, $PV = nRT$, provides a good approximation for the behavior of gases under conditions of high temperature and low pressure. However, real gases deviate from ideal behavior because ideal gas theory makes two key assumptions:

   1. Gas molecules have negligible volume compared to the total volume of the container.
   2. There are no intermolecular forces between gas molecules.

   For real gases, especially at high pressures or low temperatures, these assumptions break down. The van der Waals equation of state modifies the ideal gas law to account for these deviations:

   • Correction for intermolecular forces: Real gas molecules attract each other. This attraction reduces the force with which molecules hit the container walls, leading to a lower observed pressure than predicted by the ideal gas law. The term $\frac{an^2}{V^2}$ is added to the pressure term $P$ to account for this reduction, where $a$ is a constant specific to the gas that reflects the strength of intermolecular attractions.
   • Correction for finite molecular volume: Real gas molecules occupy a finite volume. This means the actual volume available for the molecules to move in (the "free volume") is less than the total volume of the container. The term $nb$ is subtracted from the total volume $V$, where $b$ is a constant specific to the gas that represents the volume excluded per mole of gas molecules.

   The van der Waals equation of state is:

   $(P + \frac{an^2}{V^2})(V - nb) = nRT$

   where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $T$ is temperature, $R$ is the ideal gas constant, and $a$ and $b$ are van der Waals constants specific to the gas.

   Let's apply this equation to find the pressure of ammonia gas.

   1. Identify the given values and convert units:

      • Number of moles, $n = 2$ mol
      • Volume, $V = 5.0$ L
      • Temperature, $T = 27^\circ\text{C}$. We must convert this to Kelvin: $T = 27 + 273 = 300$ K
      • van der Waals constant $a = 4.17\, \text{atm L}^2 \text{mol}^{-2}$
      • van der Waals constant $b = 0.0371\, \text{L mol}^{-1}$
      • Gas constant $R = 0.0821\, \text{L atm K}^{-1} \text{mol}^{-1}$

   2. Rearrange the van der Waals equation to solve for pressure ($P$):

      From $(P + \frac{an^2}{V^2})(V - nb) = nRT$, we can write:

      $P + \frac{an^2}{V^2} = \frac{nRT}{V - nb}$

      $P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}$

   3. Calculate the individual terms:

      • Term 1: $nRT$ …

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3. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Two liquids ‘A’ and ‘B’ form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of ‘A’ and 3 moles of ‘B’ is 550 mm Hg. At the same temperature, if one more mole of ‘B’ is added to the solution, the vapour pressure of solution increases to 560 mm Hg. Then the ratio of vapour pressures of A and B in their pure state is (A) $2:3$ (B) $3:2$ (C) $1:3$ (D) $3:1$

   ›Reveal solutionSolution

   Applying Raoult's law to the two mixtures gives $p_B^0 = 600$ and $p_A^0 = 400\text{ mm Hg}$, so $p_A^0 : p_B^0 = 2:3$. The correct option is (A).

   ---

   Concept

   For an ideal solution the total vapour pressure is the sum of the partial pressures (Raoult's law):

   $p = x_A\,p_A^0 + x_B\,p_B^0$

   Mixture 1: 1 mol A + 3 mol B ($x_A = \tfrac14,\ x_B = \tfrac34$), $p = 550\text{ mm Hg}$:

   $\frac{1}{4}p_A^0 + \frac{3}{4}p_B^0 = 550 \implies p_A^0 + 3p_B^0 = 2200 \quad (1)$

   Mixture 2: 1 mol A + 4 mol B ($x_A = \tfrac15,\ x_B = \tfrac45$), $p = 560\text{ mm Hg}$: …

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4. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Elements A and B form two non-volatile compounds (AB$_2$ and AB$_4$). When 1 g of AB$_2$ is added to 20 g of C$_6$H$_6$ (molar mass $= 78$ g mol$^{-1}$) the freezing point of C$_6$H$_6$ is lowered by 2.3 K. When 1 g of AB$_4$ is added to 20 g of C$_6$H$_6$, the freezing point of C$_6$H$_6$ was lowered by 1.3 K. The atomic masses of A and B are respectively ($K_f$(C$_6$H$_6$) $= 5.1$ Kkg mol$^{-1}$) (A) 25.59u, 42.64u (B) 42.64u, 25.59u (C) 50.29u, 31.61u (D) 31.61u, 50.29u

   ›Reveal solutionSolution

   This problem uses the colligative property of freezing point depression to determine the molar masses of two compounds, AB$_2$ and AB$_4$. By setting up two equations based on the freezing point depression formula and then relating these molar masses to the atomic masses of A and B, we find the atomic masses of A and B to be approximately 25.59 u and 42.64 u, respectively.

   The core concept here is freezing point depression, a colligative property. Colligative properties depend solely on the number of solute particles in a solution, not on their chemical identity. When a non-volatile solute is added to a solvent, it lowers the solvent's freezing point. This lowering, $\Delta T_f$, is directly proportional to the molality ($m$) of the solution.

   By measuring the freezing point depression for two different compounds (AB$_2$ and AB$_4$) dissolved in the same solvent (C$_6$H$_6$), we can determine their respective molar masses. Once we have the molar masses of AB$_2$ and AB$_4$, we can set up a system of two linear equations involving the atomic masses of A and B, and then solve for these unknown atomic masses.

   1. Recall the Freezing Point Depression Formula

      The change in freezing point, $\Delta T_f$, is given by:

      $\Delta T_f = K_f \cdot m$

      where:

      • $\Delta T_f$ is the freezing point depression (in K).
      • $K_f$ is the molal freezing point depression constant (cryoscopic constant) of the solvent (in K kg mol$^{-1}$).
      • $m$ is the molality of the solution (in mol kg$^{-1}$).

      Molality ($m$) is defined as the moles of solute per kilogram of solvent:

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

      And the moles of solute can be expressed as:

      $\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}$

      Combining these, we get:

      $m = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent (kg)}}$

   2. Calculate the Molar Mass of AB$_2$

      For the first compound, AB$_2$:

      • Mass of AB$_2 = 1$ g
      • Mass of C$_6$H$_6 = 20$ g $= 0.020$ kg
      • $\Delta T_f = 2.3$ K
      • $K_f = 5.1$ K kg mol$^{-1}$

      Substitute these values into the freezing point depression formula:

      $2.3 \text{ K} = 5.1 \text{ K kg mol}^{-1} \times \frac{1 \text{ g}}{M_{AB_2} \text{ g mol}^{-1} \times 0.020 \text{ kg}}$

      Rearranging to solve for $M_{AB_2}$:

      $M_{AB_2} = \frac{5.1 \times 1}{2.3 \times 0.020}$

      $M_{AB_2} = \frac{5.1}{0.046}$

      $M_{AB_2} \approx 110.8696 \text{ g mol}^{-1}$

   3. Calculate the Molar Mass of AB$_4$

      For the second compound, AB$_4$:

      • Mass of AB$_4 = 1$ g
      • Mass of C$_6$H$_6 = 20$ g $= 0.020$ kg
      • $\Delta T_f = 1.3$ K
      • $K_f = 5.1$ K kg mol$^{-1}$

      Substitute these values into the freezing point depression formula:

      $1.3 \text{ K} = 5.1 \text{ K kg mol}^{-1} \times \frac{1 \text{ g}}{M_{AB_4} \text{ g mol}^{-1} \times 0.020 \text{ kg}}$ …

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5. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.A 200 mL of 30% (V/V) of a solution is mixed with 500 mL of 40% (V/V) another solution. What is the volume percentage of resultant solution? (A) 33.14 (B) 35.00 (C) 26.24 (D) 37.14

   ›Reveal solutionSolution

   To find the volume percentage of the resultant solution, we calculate the total volume of solute and divide it by the total volume of the mixed solutions, then multiply by 100. The resultant solution has a volume percentage of $\boxed{37.14\%}$.

   When solutions are mixed, their concentrations don't simply average out unless their volumes are equal. The key idea is that the amount of solute from each solution combines, and the total volume of the solutions also combines. The final concentration depends on the ratio of the total solute to the total volume.

   Volume percentage (V/V) is a way to express the concentration of a solution. It is defined as the volume of solute present in 100 units of volume of the solution.

   $\text{Volume percentage (V/V)} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100\%$

   To find the volume percentage of the resultant solution, we need to:

   1. Calculate the volume of solute contributed by each initial solution.
   2. Sum these volumes to get the total volume of solute.
   3. Sum the volumes of the initial solutions to get the total volume of the resultant solution.
   4. Use the formula above with the total solute volume and total solution volume.

   Let's apply this step-by-step.

   1. Calculate the volume of solute in the first solution:

      The first solution has a volume of $200 \text{ mL}$ and is $30\%$ (V/V). This means $30 \text{ mL}$ of solute is present in every $100 \text{ mL}$ of solution.

      Volume of solute in the first solution $= 200 \text{ mL} \times \frac{30}{100} = 60 \text{ mL}$.

   2. Calculate the volume of solute in the second solution:

      The second solution has a volume of $500 \text{ mL}$ and is $40\%$ (V/V).

      Volume of solute in the second solution $= 500 \text{ mL} \times \frac{40}{100} = 200 \text{ mL}$.

   3. Calculate the total volume of solute:

      The total volume of solute in the resultant mixture is the sum of the solute volumes from both solutions.

      Total volume of solute $= 60 \text{ mL} + 200 \text{ mL} = 260 \text{ mL}$. …

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6. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.The concentration of 1 L of CaCO$_3$ solution is $10^{-5}$ M. Its concentration in ppm is (Ca = 40u; C = 12u; O = 16u) (A) 10 (B) 1000 (C) 100 (D) 1

   ›Reveal solutionSolution

   The key is to convert molarity (mol/L) to ppm (mg/L) using the molar mass of CaCO₃. A 10⁻⁵ M solution has a concentration of 1 ppm, so the correct option is (D).

   Concept & Intuition

   ppm (parts per million) for dilute aqueous solutions is essentially mg/L, because 1 L of water weighs about 1 million mg. So to go from molarity (moles per liter) to ppm, we multiply by the molar mass (g/mol) to get grams per liter, then convert grams to milligrams (×1000). That gives mg/L = ppm. Here, the molar mass of CaCO₃ is 100 g/mol, and the concentration is tiny — 10⁻⁵ mol/L — so the result will be a small number of ppm.

   Step-by-step solution

   1. Find the molar mass of CaCO₃

      Ca = 40, C = 12, O₃ = 3×16 = 48

      Total = 40 + 12 + 48 = 100 g/mol.

   2. Convert molarity to grams per liter

      Concentration = $10^{-5}$ mol/L.

      Mass per liter = $10^{-5} \text{ mol/L} \times 100 \text{ g/mol} = 10^{-3} \text{ g/L}$.

   3. Convert grams per liter to milligrams per liter (ppm)

      $1 \text{ g} = 1000 \text{ mg}$, so

      $10^{-3} \text{ g/L} = 10^{-3} \times 1000 = 1 \text{ mg/L}$. …

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7. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.10 mL of 0.5 M NaCl is required to coagulate 1L of Sb$_2$S$_3$ sol in 2 hours time. The flocculating value of NaCl (in milli moles) is (A) 20 (B) 10 (C) 5 (D) 15

   ›Reveal solutionSolution

   The flocculating value is the minimum millimoles of electrolyte per litre of sol needed to cause coagulation. Here, 10 mL of 0.5 M NaCl provides 5 millimoles of NaCl, and since that coagulates 1 L of sol, the flocculating value is 5 millimoles per litre. The correct option is (C).

   The key idea is that flocculating value is defined as the concentration (in millimoles per litre) of an electrolyte that is just sufficient to cause coagulation of a sol. It is not the total amount added, but the amount per litre of sol. So we need to find how many millimoles of NaCl are present in the volume that coagulates exactly 1 L of sol, and then express that as millimoles per litre.

   1. Find the millimoles of NaCl used.

      We have 10 mL of 0.5 M NaCl.

      Moles = Molarity × Volume (in litres) = $0.5 \times 0.010 = 0.005$ moles.

      Millimoles = $0.005 \times 1000 = 5$ millimoles.

   2. Interpret the meaning.

      These 5 millimoles of NaCl, when added to 1 L of Sb₂S₃ sol, cause coagulation in 2 hours. The flocculating value is defined as the minimum millimoles of electrolyte per litre of sol required to cause coagulation. Since exactly this amount (5 millimoles) is used for 1 L of sol, the flocculating value is simply 5 millimoles per litre.

   3. Check the options.

      (A) 20, (B) 10, (C) 5, (D) 15.

      Our computed value is 5, which matches option (C). …

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8. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
   Q.The boiling point of aqueous glucose solution is 101.3 $^\circ$C. Its freezing point is (Given that $K_f$ of water = 1.86 $^\circ$C/m and $K_b$ of water = 0.52 $^\circ$C/m) (A) 4.65 $^\circ$C (B) -4.65 $^\circ$C (C) 2.65 $^\circ$C (D) -2.65 $^\circ$C

   ›Reveal solutionSolution

   We use the given boiling point elevation to determine the molality of the glucose solution, and then use this molality to calculate the freezing point depression, finding the freezing point to be $\boxed{-4.65 \ ^\circ\text{C}}$.

   Colligative properties are those properties of solutions that depend only on the number of solute particles in the solution, not on their identity. Boiling point elevation and freezing point depression are two such properties. When a non-volatile solute like glucose is added to a solvent (water in this case), the boiling point of the solution increases, and the freezing point decreases.

   The extent of this change is directly proportional to the molality ($m$) of the solution, which is the number of moles of solute per kilogram of solvent. For non-electrolytes like glucose, the van't Hoff factor ($i$) is 1, meaning each dissolved molecule acts as a single particle.

   Here's how we can determine the freezing point:

   1. Determine the boiling point elevation ($\Delta T_b$). The normal boiling point of pure water is $100 \ ^\circ\text{C}$. The solution boils at $101.3 \ ^\circ\text{C}$. The elevation in boiling point is the difference between the solution's boiling point and the pure solvent's boiling point:

   $\Delta T_b = T_{b, \text{solution}} - T_{b, \text{solvent}}$

   $\Delta T_b = 101.3 \ ^\circ\text{C} - 100 \ ^\circ\text{C} = 1.3 \ ^\circ\text{C}$

   2. Calculate the molality ($m$) of the glucose solution. The boiling point elevation is related to the molality by the formula:

      $\Delta T_b = i \cdot K_b \cdot m$

      For glucose, a non-electrolyte, the van't Hoff factor $i=1$. We are given $K_b$ for water as $0.52 \ ^\circ\text{C/m}$.

      Substituting the values:

   $1.3 \ ^\circ\text{C} = 1 \cdot (0.52 \ ^\circ\text{C/m}) \cdot m$

   Now, solve for $m$:
   

   $m = \frac{1.3 \ ^\circ\text{C}}{0.52 \ ^\circ\text{C/m}} = 2.5 \text{ m}$

   The molality of the glucose solution is $2.5 \text{ mol/kg}$.
   

   3. Calculate the freezing point depression ($\Delta T_f$).

   The freezing point depression is related to the molality by the formula:

   > [!FORMULA]

   > $\Delta T_f = i \cdot K_f \cdot m$

   Again, for glucose, $i=1$. We are given $K_f$ for water as $1.86 \ ^\circ\text{C/m}$. …

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9. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.At T(K), copper (atomic mass = 63.5 u) has fcc structure with an edge length of x Å. The density of copper (in g cm$^{-3}$) at that temperature approximately is ($N_A = 6.0 \times 10^{23}$ mol$^{-1}$) (A) $\dfrac{423}{x}$ (B) $\dfrac{4.23}{x^3}$ (C) $\dfrac{423}{x^3}$ (D) $\dfrac{212.5}{x^3}$

   ›Reveal solutionSolution

   The density of a crystal is mass per unit volume; for an fcc unit cell, there are 4 atoms per cell, so density = (4 × atomic mass) / (Avogadro’s number × edge length³). Substituting values gives density ≈ 423 / x³ g cm⁻³, matching option (C).

   Concept & Intuition

   Density of a crystalline solid is simply the mass of one unit cell divided by its volume. For a face-centered cubic (fcc) lattice, each unit cell contains 4 atoms (not 1 — a common mistake). The mass of those 4 atoms comes from the atomic mass and Avogadro’s number. The edge length is given in Å, but we need the volume in cm³, so we convert Å to cm (1 Å = 10⁻⁸ cm). The problem gives numbers that are deliberately rounded to make the arithmetic clean.

   Step-by-step solution

   1. Number of atoms per fcc unit cell

      In fcc, atoms are at the 8 corners (each shared by 8 cells → 1/8 atom per corner) and at the 6 face centers (each shared by 2 cells → 1/2 atom per face).

      Total = $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4$ atoms per unit cell.

   2. Mass of one unit cell

      Atomic mass of copper = 63.5 g mol⁻¹.

      Mass of one atom = $\frac{63.5}{N_A}$ grams, with $N_A = 6.0 \times 10^{23}$.

      So mass of 4 atoms = $4 \times \frac{63.5}{6.0 \times 10^{23}} = \frac{254}{6.0 \times 10^{23}} = \frac{42.33}{10^{23}}$ g (approximately).

   3. Volume of the unit cell

      Edge length = $x$ Å = $x \times 10^{-8}$ cm.

      Volume = $(x \times 10^{-8})^3 = x^3 \times 10^{-24}$ cm³.

   4. Density formula

   $\rho = \frac{\text{mass of cell}}{\text{volume of cell}} = \frac{\frac{254}{6.0 \times 10^{23}}}{x^3 \times 10^{-24}} = \frac{254}{6.0 \times 10^{23}} \times \frac{1}{x^3 \times 10^{-24}}$

   Simplify the powers of ten: $\frac{1}{10^{23} \times 10^{-24}} = \frac{1}{10^{-1}} = 10$. …

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10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.The vapour pressure of a pure liquid A is 70 torr at 300 K. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and total vapour pressure of the solution is 84 torr at the same temperature. The vapour pressure of pure liquid B (in torr) is (A) 140 (B) 90 (C) 120 (D) 80

    ›Reveal solutionSolution

    Using Raoult’s law for an ideal solution, the total vapour pressure is the sum of the partial pressures: $P = x_A P_A^0 + x_B P_B^0$. With $P = 84$ torr, $x_B = 0.2$, $x_A = 0.8$, and $P_A^0 = 70$ torr, solving gives $P_B^0 = 140$ torr.

    Concept & Intuition

    Raoult’s law says that in an ideal solution, each component’s partial vapour pressure is its mole fraction times its pure vapour pressure. The total pressure is just the sum of these. Here, we know the total, one pure pressure, and one mole fraction — so we can solve for the unknown pure pressure. The key is to remember that mole fractions of A and B add to 1.

    Step-by-step solution

    1. Identify the given data

       • Pure vapour pressure of A: $P_A^0 = 70$ torr
       • Mole fraction of B: $x_B = 0.2$
       • Total vapour pressure of solution: $P = 84$ torr
       • Temperature is constant at 300 K.

    2. Find the mole fraction of A

       Since only A and B are present:

    $x_A + x_B = 1 \quad \Rightarrow \quad x_A = 1 - 0.2 = 0.8$

    3. Apply Raoult’s law for each component

       Partial pressure of A: $p_A = x_A P_A^0 = 0.8 \times 70 = 56$ torr

       Partial pressure of B: $p_B = x_B P_B^0 = 0.2 \times P_B^0$

    4. Total pressure is the sum of partial pressures

       $P = p_A + p_B = 56 + 0.2 P_B^0 = 84$ …

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11. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.An aqueous solution of a non-volatile solute boils at $100.17\,^\circ\mathrm{C}$. The temperature at which this solution will freeze (in $^\circ\mathrm{C}$) is $(K_b(\mathrm{H}_2\mathrm{O}) = 0.512\,^\circ\mathrm{C}\,\mathrm{kg}\,\mathrm{mol}^{-1},\ K_f(\mathrm{H}_2\mathrm{O}) = 1.86\,^\circ\mathrm{C}\,\mathrm{kg}\,\mathrm{mol}^{-1})$ (A) $-0.62$ (B) $-0.512$ (C) $-1.24$ (D) $-1.86$

    ›Reveal solutionSolution

    Using the boiling point elevation to find the molality, then applying freezing point depression with that same molality gives a freezing point of $-0.62\,^\circ\mathrm{C}$, so the correct option is (A).

    Concept & Intuition

    Both boiling point elevation and freezing point depression are colligative properties — they depend only on the number of solute particles, not on their identity. Since the same solution is used, the molality is the same for both calculations. The key is to first extract the molality from the given boiling point data, then plug it into the freezing point formula. The ratio of the two constants ($K_f / K_b$) gives a direct shortcut.

    Step-by-step solution

    1. Boiling point elevation The normal boiling point of pure water is $100.00\,^\circ\mathrm{C}$. The observed boiling point is $100.17\,^\circ\mathrm{C}$, so the elevation is

    $\Delta T_b = 100.17 - 100.00 = 0.17\,^\circ\mathrm{C}.$

    The formula is $\Delta T_b = K_b \cdot m$, where $m$ is the molality. Thus

    $m = \frac{\Delta T_b}{K_b} = \frac{0.17}{0.512}.$

    2. Freezing point depression For water, the normal freezing point is $0.00\,^\circ\mathrm{C}$. The depression is $\Delta T_f = K_f \cdot m$. Substituting the molality from step 1:

    $\Delta T_f = K_f \cdot \frac{\Delta T_b}{K_b} = 1.86 \times \frac{0.17}{0.512}.$

    3. Calculate numerically First compute $\frac{0.17}{0.512} \approx 0.33203$. Then $$$$

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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.On passing a current of $1.2~\mathrm{A}$ through a solution of salt of copper for $40~\mathrm{min}$, $0.96~\mathrm{g}$ of copper was deposited. The equivalent weight of copper in g is (A) 21.2 (B) 31.75 (C) 63.5 (D) 15.9

    ›Reveal solutionSolution

    We use Faraday's First Law of Electrolysis, which states that the mass of a substance deposited is directly proportional to the charge passed. By calculating the total charge and applying the formula, we find the equivalent weight of copper to be approximately $32.17~\mathrm{g}$. The closest option, and the standard equivalent weight for $\mathrm{Cu^{2+}}$, is $\boxed{31.75}$.

    Concept and Intuition

    Electrolysis is the process of using electrical energy to drive non-spontaneous chemical reactions. When an electric current is passed through a solution of a metal salt, metal ions migrate to the cathode (negative electrode) and gain electrons, getting reduced and deposited as neutral metal atoms.

    The fundamental principle governing the amount of substance deposited during electrolysis is Faraday's First Law of Electrolysis. This law states that the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte.

    Mathematically, this can be expressed as:

    $m \propto Q$

    where $m$ is the mass of the substance deposited and $Q$ is the total charge passed.

    To convert this proportionality into an equation, we introduce the concept of equivalent weight ($E$). The equivalent weight of a substance is the mass of that substance deposited or liberated by one Faraday of charge.

    Important

    One Faraday ($F$) is the charge carried by one mole of electrons, which is approximately $96500~\mathrm{C/mol}$ (Coulombs per mole of electrons). It can also be thought of as $96500~\mathrm{C/equivalent}$.

    The relationship between mass deposited ($m$), equivalent weight ($E$), and charge ($Q$) is given by:

    $m = \frac{E}{F} \times Q$

    The total charge ($Q$) passed through the solution can be calculated from the current ($I$) and time ($t$) using the formula:

    $Q = I \times t$

    Where $Q$ is in Coulombs ($\mathrm{C}$), $I$ is in Amperes ($\mathrm{A}$), and $t$ is in seconds ($\mathrm{s}$).

    Our goal is to find the equivalent weight ($E$) of copper using the given experimental data. We will first calculate the total charge passed and then use Faraday's First Law to determine $E$.

    Step-by-step Solution

    1. Convert time to seconds:

       The given time is $40~\mathrm{min}$. To use the formula $Q = I \times t$ with current in Amperes, time must be in seconds.

       $t = 40~\mathrm{min} \times 60~\mathrm{s/min} = 2400~\mathrm{s}$

    2. Calculate the total charge passed ($Q$):

       The current ($I$) is $1.2~\mathrm{A}$ and the time ($t$) is $2400~\mathrm{s}$.

       $Q = I \times t$

       $Q = 1.2~\mathrm{A} \times 2400~\mathrm{s}$

       $Q = 2880~\mathrm{C}$

    3. Apply Faraday's First Law to find the equivalent weight ($E$): …

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