The depression in the freezing point observed for a formic acid solution of concentration 0.5 mL L–1 is 0.0405°C. Density of formic acid is 1.05 g mL–1. The Van’t Hoff factor of the formic acid solution is nearly (Given for water kf = 1.86 k kg mol–1)
The depression in the freezing point observed for a formic acid solution of concentration 0.5 mL L–1 is 0.0405°C. Density of formic acid is 1.05 g mL–1. The Van’t Hoff factor of the formic acid solution is nearly (Given for water kf = 1.86 k kg mol–1)
- 0.8
- 1.1
- 1.9
- 2.4
The Correct Option is C
Solution and Explanation
ΔTf of formic acid = 0.0405°C
Concentration = 0.5 mL/L
and density = 1.05 g/mL
∴ Mass of formic acid in solution = 1.05 × 0.5 g
= 0.525 g
∴ According to Van’t Hoff equation,
ΔT_f=ik_f.m
0.0405=\(\frac{i×1.86×0.525}{46×1}\)
(Assuming mass of 1 L water = kg)
i=\(\frac{0.0405×46}{1.86×0.525}\)=1.9
Top Questions on Solutions
- Arrange the following in increasing order of solubility product: \[ {Ca(OH)}_2, {AgBr}, {PbS}, {HgS} \]
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is \(\_\_\_\_\_\) x \(10^{-2}\)M. (Nearest integer)
- 0.1 M solution of KI reacts with excess of \( \text{H}_2\text{SO}_4 \) and KIO\(_3\), according to the equation: \[ 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O \] Identify the correct statements:
Questions Asked in JEE Main exam
- If \( \sin x + \sin^2 x = 1 \), \( x \in \left(0, \frac{\pi}{2} \right) \), then the expression \[ (\cos^2 x + \tan^2 x) + 3(\cos^4 x + \tan^4 x + \cos^4 x + \tan^4 x) + (\cos^6 x + \tan^6 x) \] is equal to:
- JEE Main - 2025
- Trigonometric Identities
- The sum of the squares of all the roots of the equation \( x^2 + [2x - 3] - 4 = 0 \) is:
- JEE Main - 2025
- Quadratic Equations
- Let ABCD be a trapezium whose vertices lie on the parabola \( y^2 = 4x \). Let the sides AD and BC of the trapezium be parallel to the y-axis. If the diagonal AC is of length \( \frac{25}{4} \) and it passes through the point \( (1, 0) \), then the area of ABCD is:
- JEE Main - 2025
- Coordinate Geometry
- A molecule (P) on treatment with acid undergoes rearrangement and gives (Q). (Q) on ozonolysis followed by reflux under alkaline condition gives (R). The structure of (R) is given below. The structure of (P) is:
- JEE Main - 2025
- Organic Chemistry
- The product B formed in the following reaction sequence is: \[ {C}_6{H}_5{CN} \xrightarrow{{HCl}} (A) \xrightarrow{{AgCN}} (B) \]
- JEE Main - 2025
- Organic Chemistry
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.