Problem based on lowering in vapour pressure
Help me to solve this problem. An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is: (a) 0.02 (b)1 (c) 20 (d)180
Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M
The contribution of an electrolyte, or an ion electrolyte, is reported as the molar of a conductance. The definition of the molar conductance is based on the following conductivity cell in which the electrodes are 1 m apart and of sufficient area that th
Avogadro's hypothesis Law Principle- Berzelius, a chemist tried
Briefly describe the procedure for separating the components of the gun-powder?
The amount of KMnO4 required to prepare 100 ml of 0.1N solution in alkaline medium is: (a) 1.58 gm (b) 3.16 gm (c) 0.52 gm (d) 0.31 gmAnswer: (a) In alkaline medium KMnO4 act as ox
The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter
The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A
20 g of hydrogen is present in 5 litre of vessel. Determine he molar concentration of hydrogen: (a) 4 (b) 1 (c) 3 (d) 2 Choose the right answer from above.
Which of the following would have the maximum osmotic pressure (assume that all salts are 90% dissociated): (a) Decimolar aluminium sulphate (b) Decimolar barium chloride (c) Decimolar sodium sulphate (d) A solution obtained by mix
20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is: (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9 Choose the right answer from above.
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