Question:
Which of the following statements are correct ?
A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point
B. Aniline can be purified by steam distillation as aniline is miscible in water.
C. Ethanol can be separated from ethanol water mixture by azeotropic distillation because it forms azeotrope.
D. An organic compound is pure, if mixed M.P. is remained same.
Choose the most appropriate answer from the options given below :
Which of the following statements are correct ?
A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point
B. Aniline can be purified by steam distillation as aniline is miscible in water.
C. Ethanol can be separated from ethanol water mixture by azeotropic distillation because it forms azeotrope.
D. An organic compound is pure, if mixed M.P. is remained same.
Choose the most appropriate answer from the options given below :
A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point
B. Aniline can be purified by steam distillation as aniline is miscible in water.
C. Ethanol can be separated from ethanol water mixture by azeotropic distillation because it forms azeotrope.
D. An organic compound is pure, if mixed M.P. is remained same.
Choose the most appropriate answer from the options given below :
Updated On: Mar 21, 2025
- A, B, C only
- A, C, D only
- B, C, D only
- A, B, D only
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
• A: True. Glycerol decomposes at its normal boiling point, so it is purified by vacuum distillation.
• B: False. Aniline is immiscible in water but can still be purified by steam distillation.
• C: True. Ethanol forms an azeotrope with water, making azeotropic distillation effective for separation.
• D: True. A pure organic compound exhibits the same melting point even when mixed with another sample of the same compound.
Was this answer helpful?
0
0
Top Questions on Organic Chemistry
- Given below are two statements: Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas. Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:
- JEE Main - 2025
- Chemistry
- Organic Chemistry
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
- JEE Main - 2025
- Chemistry
- Organic Chemistry
- Identify A and B in each of the following reaction sequence:
(a) \[ CH_3CH_2Cl \xrightarrow{NaCN} A \xrightarrow{H_2/Ni} B \]
(b) \[ C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} A \xrightarrow{C_6H_5NH_2} B \]- CBSE CLASS XII - 2025
- Chemistry
- Organic Chemistry
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
- CBSE Class X - 2025
- Science
- Organic Chemistry
- Carbohydrates are polyhydroxy aldehydes or ketones that represent enormous structural diversity in terms of the arrangement of atoms in space, resulting in hundreds of stereoisomers. Although the chemical properties of most stereoisomers may not be very different, their metabolic rate and utilization in biological systems is significantly different and known to influence the overall carbohydrate metabolism. Structural variants, which arise due to a different arrangement of atoms in three-dimensional space are known as stereoisomers. The number of stereoisomers can be theoretically estimated by using the formula 2n, where ‘n’ is the number of stereocenters or asymmetric (chiral) carbon atoms in a molecule. Out of these stereoisomers, there are some structures, which are mirror images of each other, and they are referred to as enantiomers.
- CBSE CLASS XII - 2025
- Chemistry
- Organic Chemistry
View More Questions
Questions Asked in JEE Main exam
- Consider a long thin conducting wire carrying a uniform current \( I \). A particle having mass \( M \) and charge \( q \) is released at a distance \( a \) from the wire with a speed \( v_0 \) along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance \( x \) from the wire. The value of \( x \) is:
- JEE Main - 2025
- Magnetic Field
- Two equal sides of an isosceles triangle are along \( -x + 2y = 4 \) and \( x + y = 4 \). If \( m \) is the slope of its third side, then the sum of all possible distinct values of \( m \) is:
- JEE Main - 2025
- 3D Geometry
- Let the position vectors of the vertices A, B, and C of a tetrahedron ABCD be \( \hat{i} + 2\hat{j} + \hat{k} \), \( \hat{i} + 3\hat{j} - 2\hat{k} \), and \( 2\hat{i} + \hat{j} - \hat{k} \) respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is \( \frac{\sqrt{10}}{3} \) and the volume of the tetrahedron is \( \frac{\sqrt{805}}{6\sqrt{2}} \), then the position vector of E is:
- JEE Main - 2025
- Geometry and Vectors
- If \( x_1, x_2, x_3, x_4 \) are in GP (Geometric Progression), then we subtract 2, 4, 7, and 8 from \( x_1, x_2, x_3, x_4 \) respectively, then the resultant numbers are in AP (Arithmetic Progression). Then the value of \( \frac{1}{24} (x_1 \cdot x_2 \cdot x_3 \cdot x_4) \) is:
- JEE Main - 2025
- Arithmetic Progression
- Let the area of a triangle \( \triangle PQR \) with vertices \( P(5, 4) \), \( Q(-2, 4) \), and \( R(a, b) \) be 35 square units. If its orthocenter and centroid are \( O(2, \frac{14}{5}) \) and \( C(c, d) \) respectively, then \( c + 2d \) is equal to:
- JEE Main - 2025
- Geometry and Vectors
View More Questions