At $27^{\circ} C$, a solution containing $25 \,g$ of solute in $2500 \,mL$ of solution exerts an osmotic pressure of $400 \,Pa$. The molar mass of the solute is ____ $g\,mol ^{-1}$. (Nearest integer)
(Given : $R =0.083 \,L\,bar\, K ^{-1} \,mol ^{-1}$ )
At $27^{\circ} C$, a solution containing $25 \,g$ of solute in $2500 \,mL$ of solution exerts an osmotic pressure of $400 \,Pa$. The molar mass of the solute is ____ $g\,mol ^{-1}$. (Nearest integer)
(Given : $R =0.083 \,L\,bar\, K ^{-1} \,mol ^{-1}$ )
Show Hint
Correct Answer: 62250
Approach Solution - 1
\(\pi=CRT \)
\(\frac{400Pa}{10^5}=\frac{\frac{2.5g}{M_o}}{\frac{250}{1000L}}\times0.83\frac{L−bar}{K⋅mol}\times300K \)
\(M_o=62250\)
So, the answer is 62250.
Approach Solution -2
The osmotic pressure (\(\pi\)) of a solution is given by the formula:
\(\pi = MRT\)
where M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin.
To find the molar mass of the solute, we need to first find the molarity of the solution:
\(\text{Molarity }(M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)
We are given that the solution contains 2.5g of solute in 250.0mL of solution. To convert mL to L, we divide by 1000:
\(\text{Volume of solution} = \frac{250.0 mL}{1000} = 0.250 L\)
To find the moles of solute, we need to use the formula:
\(\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass}}\)
Let's assume that the molar mass of the solute is x g/mol. Then we have:
\(\text{moles of solute} = \frac{2.5 g}{x} g/mol\)
Now we can calculate the molarity of the solution:
\(\text{Molarity }(M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)
\(M = \frac{\frac{2.5 g}{x g/mol}}{0.250 L}\)
\(M = \frac{10}{x} mol/L\)
Next, we can use the given osmotic pressure and temperature to find the molarity of the solution:
\(\pi = MRT\)
\(400 Pa = \frac{10}{x} mol/L \times 0.083 L bar K^{-1} mol^{-1} \times (27°C + 273.15) K\)
Simplifying and solving for x, we get:
\(x = 165\ gmol^{-1}\) (nearest integer)
Therefore, the molar mass of the solute is \(165 \ gmol^{-1}\), i.e., 62250.
Top Questions on Colligative Properties
- Which among the following has the highest molar elevation constant?
- KEAM - 2025
- Chemistry
- Colligative Properties
- The percentage dissociation of a salt (MX$_3$) solution at a given temperature (van't Hoff factor $i = 2$) is .................. % (Nearest integer)
- JEE Main - 2025
- Chemistry
- Colligative Properties
Given below are two statements:
Statement (I): Molal depression constant $ k_f $ is given by $ \frac{M_1 R T_f}{\Delta S_{\text{fus}}} $, where symbols have their usual meaning.
Statement (II): $ k_f $ for benzene is less than the $ k_f $ for water.
In light of the above statements, choose the most appropriate answer from the options given below:- JEE Main - 2025
- Chemistry
- Colligative Properties
- 180 g of glucose, \(C_6H_{12}O_6\), is dissolved in 1 kg of water in a vessel. The temperature at which water boils at 1.013 bar is ______ (given, \(K_b\) for water is 0.52 K kg mol\(^{-1}\)). Boiling point for pure water is 373.15 K
- KCET - 2025
- Chemistry
- Colligative Properties
- Which of the following solutions will show highest freezing point depression?
- MHT CET - 2025
- Chemistry
- Colligative Properties
Questions Asked in JEE Main exam
- Let \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to __________.
- JEE Main - 2025
- Differential equations
- Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is:
- JEE Main - 2025
- Statistics
- Let the system of equations $ x + 5y - z = 1 $ $ 4x + 3y - 3z = 7 $ $ 24x + y + \lambda z = \mu $ where $ \lambda, \mu \in \mathbb{R} $, have infinitely many solutions. Then the number of the solutions of this system, if $x, y, z$ are integers and satisfy $7 \leq x + y + z \leq 77$, is:
- JEE Main - 2025
- Linear Systems and Determinants
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
Match List-I with List-II: List-I
- JEE Main - 2025
- Classification Of Organic Compounds
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.