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
- CrCl\(_3\).xNH\(_3\) can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558°C. Assuming 100\% ionization of this complex and coordination number of Cr is 6, the complex will be:
- JEE Main - 2025
- Chemistry
- Colligative Properties
- Given below are two statements:
Statement (I): NaCl is added to the ice at \( 0^\circ C \), present in the ice cream box to prevent the melting of ice cream.
Statement (II): On addition of NaCl to ice at \( 0^\circ C \), there is a depression in freezing point.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Chemistry
- Colligative Properties
- Consider the given plots of vapor pressure (VP) vs temperature (T/K). Which amongst the following options is the correct graphical representation showing \( \Delta T_f \) depression in the freezing point of solvent in a solution?
- JEE Main - 2025
- Chemistry
- Colligative Properties
If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
- JEE Main - 2025
- Chemistry
- Colligative Properties
1.24 g of \(AX_2\) (molar mass 124 g mol\(^{-1}\)) is dissolved in 1 kg of water to form a solution with boiling point of 100.105°C, while 2.54 g of AY_2 (molar mass 250 g mol\(^{-1}\)) in 2 kg of water constitutes a solution with a boiling point of 100.026°C. \(Kb(H)_2\)\(\text(O)\) = 0.52 K kg mol\(^{-1}\). Which of the following is correct?
- JEE Main - 2025
- Chemistry
- Colligative Properties
Questions Asked in JEE Main exam
- In YDSE, light of intensity \( 4I \) and \( 9I \) passes through two slits respectively. Difference of maximum and minimum intensity of interference pattern is:
- JEE Main - 2025
- Wave optics
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
- JEE Main - 2025
- Differential Equations
Find the IUPAC name of the compound.
- JEE Main - 2025
- Aromaticity & chemistry of aromatic compounds
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
- JEE Main - 2025
- Limits and Exponential Functions
- The integral \[ \int_0^\pi \frac{8x}{4\cos^2 x + \sin^2 x} \, dx \text{ is equal to:} \]
- JEE Main - 2025
- Trigonometric Functions
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.