We say in a solution made up of a solvent and a solute, the majour component is solvent and the sbstance present in small propotions is solute.
But if we take the case of a 50-50 solution where each component is present at equaly, how can we diffrentiate between a solvent and a solute?
Asked by Prem | 15th May, 2017, 09:10: PM
A solution is a combination of two or more substances to form a homogeneous mixture.
Generally, substance present in more quantity is called as solvent and the other present in less quantity is called solute.
Water is a common solvent as the medium that dissolves the other substance is called the solvent; the substance that becomes dissolved is called the solute.
If the solute and solvent exist in equal quantities, such as a 50:50 mixture of ethanol and water, the concepts of "solute" and "solvent" become less relevant. In such a case, the substance that is more often used as a solvent may be designated as the solvent. In the given example, it would be water.
A solution is a combination of two or more substances to form a homogeneous mixture.
Generally, substance present in more quantity is called as solvent and the other present in less quantity is called solute.
Water is a common solvent as the medium that dissolves the other substance is called the solvent; the substance that becomes dissolved is called the solute.
If the solute and solvent exist in equal quantities, such as a 50:50 mixture of ethanol and water, the concepts of "solute" and "solvent" become less relevant. In such a case, the substance that is more often used as a solvent may be designated as the solvent. In the given example, it would be water.
Answered by Prachi Sawant | 16th May, 2017, 10:43: AM
Related Videos
- HOW TO FIND VANT OF FACTOR
- Solution
- Value of Van't Hoff factor if CH3COOH 60% dissociates and 40% dimerized is–(1) 1.2, (2) 1.4, (3) 1.6, (4) 1.8
- I know that answer is (a) but please explain the reason in detail
- Calculate the freezing point of a solution containing 0.52 g glucose (C6H12O6) dissolved in 80.20g water. For water Kf = 1.86 K Kg mol-1
- 0.90 g of an 1:2 electrolyte was dissolved in 87.90 g of benzene. This raised the boiling point of benzene by 0.250 C. If the molecular mass of the electrolyte is 103.0molar calculate the molal elevation constant for benzene. Given that the solute dissociate with degree of dissociation 0.25.
- 0.216molal solution of Cadmium sulphate is prepared in 1000gram of water. The depression in freezing point was measured to be 0.284K. Calculate the Vant hoff factor. The Cryoscopic constant for water is 1.86K Kg mol-1.
- Two molar solution of potassium Ferro cyanide has degree of dissociation 0.70 at 300K.Find out the osmotic pressure of the solution.
- substance X forms trimer when added to Benzene calculates the freezing point of the 0.25 molal solution. The degree of association of the solute is found to be 0.80. The freezing point of the benzene is 5.5oC and its Cryoscopic constant is 5.12 K m-1.
- Calculate the amount of KCl must be added to one kilogram of water so as to reduce its freezing point by two Kelvin. For water Kf = 1.86 K Kg mol-1.
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change