0.05M CuSO$_4$ when treated with 0.01M K$_2$Cr$_2$O$_7$ gives green colour solution of Cu$_2$Cr$_2$O$_7$. The \[ \text{[SPM : Semi Permeable Membrane]} \]\(K_2Cr_2O_7\) \(CuSO_4\) Side X SPM Side Y
Due to osmosis:
| \(K_2Cr_2O_7\) | \(CuSO_4\) | |
| Side X | SPM | Side Y |
Due to osmosis:
- Green colour formation observed on side Y.
- Green colour formation observed on side X.
- Molarity of\(K_2Cr_2O_7\)solution is lowered.
- Molarity of\(CuSO_4\)solution is lowered.
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to understand the concept of osmosis and how it affects the molarity of solutions separated by a semipermeable membrane (SPM).
In the given scenario, we have the following setup:
| \(K_2Cr_2O_7\) | \(CuSO_4\) | |
| Side X | SPM | Side Y |
In this arrangement, \(K_2Cr_2O_7\) is on Side X and \(CuSO_4\) is on Side Y. Osmosis refers to the movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.
Now, let's analyze the options:
- Green colour formation observed on side Y.
- Green colour formation observed on side X.
- Molarity of \(K_2Cr_2O_7\) solution is lowered.
- Molarity of \(CuSO_4\) solution is lowered.
Since the solutions are separated by a semipermeable membrane, only the solvent (typically water) can move to equalize solute concentrations, not the solute itself. Therefore, any change in molarity due to osmosis would typically be a result of water moving from the side with lower solute concentration to the side with higher solute concentration, making the latter solution more dilute.
In this situation, osmosis will cause the water to move from the \(K_2Cr_2O_7\) side (Side X) to the \(CuSO_4\) side (Side Y) because the \(CuSO_4\) solution is involved in forming the green complex \(Cu_2Cr_2O_7\), thereby reducing its effective concentration.
Thus, the molarity of the \(CuSO_4\) solution is lowered as water moves into Side Y to equalize concentration due to osmosis.
Therefore, the correct option is:
Molarity of \(CuSO_4\) solution is lowered.
Approach Solution -2
Due to osmosis, the solvent molecules from side Y (CuSO4) diffuse across the semi-permeable membrane (SPM) into side X (K2Cr2O7). This dilution effect causes the molarity of the CuSO4 solution (side Y) to decrease over time.
- CuSO4 reacts with K2Cr2O7, forming Cu2Cr2O7, which is responsible for the green color.
- The green color is observed due to the formation of Cu2Cr2O7, but the key change is in the molarity of the CuSO4 solution on side Y, which decreases as the solvent diffuses.
Thus the correct answer is Option 4.
Top JEE Main Chemistry Questions
\(t_{100\%}\) is the time required for 100% completion of a reaction, while \(t_{1/2}\) is the time required for 50% completion of the reaction. Which of the following correctly represents the relation between \(t_{100\%}\) and \(t_{1/2}\) for zero order and first order reactions respectively
- JEE Main - 2026
- Chemistry
- Zero-order Reactions
One mole of an alkane (\(x\)) requires 8 mole oxygen for complete combustion. Sum of number of carbon and hydrogen atoms in the alkane (\(x\)) is ______.
- JEE Main - 2026
- Chemistry
- Hydrocarbons
For reaction \(A \rightarrow P\), rate constant \(k = 1.5 \times 10^3\ s^{-1}\) at \(27^\circ C\). If activation energy for the above reaction is \(60\ kJ\ mol^{-1}\), then the temperature (in \(^{\circ}C\)) at which rate constant \(k = 4.5 \times 10^3\ s^{-1}\) is ______. (Nearest integer) \[ \text{Given: } \log 2 = 0.30,\ \log 3 = 0.48,\ R = 8.3\ J\ K^{-1}\ mol^{-1},\ \ln 10 = 2.3 \]
- JEE Main - 2026
- Chemistry
- Activation energy
At the transition temperature \(T\), \(A \rightleftharpoons B\) and \(\Delta G^\circ = 105 - 35\log T\), where \(A\) and \(B\) are two states of substance \(X\). The transition temperature in \(^{\circ}C\) when pressure is 1 atm is ______.
- JEE Main - 2026
- Chemistry
- Gibbs Free Energy
Identify compounds A and E in the following reaction sequence.
- JEE Main - 2026
- Chemistry
- Aromaticity & chemistry of aromatic compounds
Top JEE Main Colligative Properties Questions
- When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $x \%$ of vapor pressure of pure solvent. What is $x \%$?
- JEE Main - 2026
- Chemistry
- Colligative Properties
- ‘x’ be the osmotic pressure of a solution formed by dissolving 1g of a protein (M = 50,000 g/mol) in 0.5 litre and ‘y’ be the osmotic pressure of solution formed by dissolving 2g of the same protein in 1 litre at 300 K. If ‘z’ be the osmotic pressure of solution formed by mixing above two solutions. Then the value of ‘x’, ‘y’ and ‘z’ respectively are.
\text{Use: \( R = 0.083 \, \text{lit-bar/K-mol} \)}
- JEE Main - 2026
- Chemistry
- Colligative Properties
Molarity of \( H_2SO_4 (aq.) \) solution is 4.9 M. If the density of the solution is 1.40 g/mL, then molality and mole fraction of solute in the solution is:
(Molar mass of \( H_2SO_4 = 98 \, \text{g mol}^{-1} \))
- JEE Main - 2026
- Chemistry
- Colligative Properties
- $W$ g of a non-volatile electrolyte solid solute of molar mass $M\,\text{g mol}^{-1}$ when dissolved in $100\,\text{mL}$ water decreases vapour pressure of water from $640\,\text{mm Hg}$ to $600\,\text{mm Hg}$. If aqueous solution of the electrolyte boils at $375\,\text{K}$ and $K_b$ for water is $0.52\,\text{K kg mol}^{-1}$, then the mole fraction of the electrolyte $(x_2)$ in the solution can be expressed as ___.
(Given: density of water $=1\,\text{g mL}^{-1}$, boiling point of water $=373\,\text{K}$)
- JEE Main - 2026
- Chemistry
- Colligative Properties
- A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute A of molar mass 60 g mol$^{-1}$ and 0.9 g of a non-volatile non-electrolyte solute B of molar mass 180 g mol$^{-1}$ in 100 mL H$_2$O at 27$^\circ$C. Osmotic pressure of the solution will be
[Given: R = 0.082 L atm K$^{-1}$ mol$^{-1}$]
- JEE Main - 2026
- Chemistry
- Colligative Properties
Top JEE Main Questions
- 1 mole of ideal diatomic gas is enclosed in a cylinder piston arrangement having cross-sectional area of piston \(4\,\text{cm}^2\). If gas has only rotational modes and \(P_{atm}=100\ \text{kPa}\), some amount of heat is added to the system as a result piston moves up slowly by \(2.5\,\text{cm}\). If temperature change is \(1.2^\circ C\). Find heat given to gas.
- JEE Main - 2026
- Thermal Physics
- Let $S = \{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3} \sin\theta\}$. Then $\sum_{\theta \in S} \theta$ is equal to:
- JEE Main - 2026
- Trigonometry
Refer the figure below. \( \mu_1 \) and \( \mu_2 \) are refractive indices of air and lens material respectively. The height of image will be _____ cm.
- JEE Main - 2026
- Optics and Refraction
In single slit diffraction pattern, the wavelength of light used is \(628\) nm and slit width is \(0.2\) mm. The angular width of central maximum is \(\alpha \times 10^{-2}\) degrees. The value of \(\alpha\) is ____.
- JEE Main - 2026
- Interference and Diffraction
\(t_{100\%}\) is the time required for 100% completion of a reaction, while \(t_{1/2}\) is the time required for 50% completion of the reaction. Which of the following correctly represents the relation between \(t_{100\%}\) and \(t_{1/2}\) for zero order and first order reactions respectively
- JEE Main - 2026
- Zero-order Reactions