Boiling point of a 2% aqueous solution of a non-volatile solute A is equal to the boiling point of 8% aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is
- \(M_A = 4M_B\)
- \(M_B = 4M_A\)
- \(M_A = 8M_B\)
- \(M_B = 8M_A\)
The Correct Option is B
Solution and Explanation
\((ΔT_b)_A = (ΔT_b)_B\)
\(K_bM_A = K_bM_B\)
\(⇒ M_A = M_B\)
\(⇒ \frac{\frac{2}{M_A}}{100}×1000 = \frac{\frac{8}{M_B}}{100} × 1000\)
\(⇒ M_B = 4M_A\)
The correct answer is (B): \(M_B = 4M_A\)
Top Questions on Solutions
- Consider the dissociation equilibrium of the following weak acid: \[ \mathrm{HA \rightleftharpoons H^+(aq) + A^-(aq)} \] If the \(pK_a\) of the acid is \(4\), then the pH of a \(10\ \text{mM}\) HA solution is ________ (Nearest integer). (Given: The degree of dissociation can be neglected with respect to unity)
- The crystal field splitting energy of $[Co(oxalate)_3]^3-$ complex is 'n' times that of the $[Cr(oxalate)_3]^3-$ complex. Here 'n' is_______ (Assume $\Delta_0 > P$)}
- The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is ____________ g L$^{-1}$. (Nearest integer)
Given: R = 0.08 L atm K$^{-1}$ mol$^{-1}$
Assume complete dissociation of NaCl
(Given : Molar mass of Na and Cl are 23 and 35.5 g mol$^{-1}$ respectively.) A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.- At \(T\) K, \(100\,\text{g}\) of \(98%\) \(H_2SO_4\) (w/w) aqueous solution is mixed with \(100\,\text{g}\) of \(49%\) \(H_2SO_4\) (w/w) aqueous solution. What is the mole fraction of \(H_2SO_4\) in the resultant solution? (Given: Atomic mass \(H = 1\,u,\; S = 32\,u,\; O = 16\,u\). Assume that temperature after mixing remains constant.)
Questions Asked in JEE Main exam
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\).
- JEE Main - 2026
- Current electricity
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
- JEE Main - 2026
- p -Block Elements
- Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are \(2.4 \times 10^8\) m/s and \(2.7 \times 10^8\) m/s, respectively, then the value of critical angle is :
- JEE Main - 2026
- Newtons Laws of Motion
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is:
- JEE Main - 2026
- Thermodynamics
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
- JEE Main - 2026
- Coordinate Geometry
Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.