Determining highest normality
What is the correct answer. Which of the given solutions contains highest normality: (i) 8 gm of KOH/litre (ii) N phosphoric acid (iii) 6 gm of NaOH /100 ml (iv) 0.5M H2SO4
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
Choose the right answer from following. The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be: (a) 18.0 (b) 342 (c) 60 (d) 180
A) Two compartments each of 1 m3 capacity are joined by a valve and insulated from the surroundings and from one another. One compartment has saturated steam at 683.6 kPa and the other contains steam at the same temperature but at a pressure of 101.3 kPa. T
Give me answer of this question. What weight of ferrous ammonium sulphate is requiored to prepare 100 ml of 0.1 normal solution (mol. wt. 392): (a) 39.2 gm (b) 3.92 gm (c)1.96 gm (d)19.6 gm
The Debye Huckel theory shows how the potential energy of an ion in solution depends on the ionic strength of the solution.Except at infinite dilution, electrostatic interaction between ions alters the properties of the solution from those excepted from th
give atleast two application of following colligative properties
A mixture has 18 g water and 414 g ethanol. What is the mole fraction of water in mixture (suppose ideal behaviour of mixture): (i) 0.1 (ii) 0.4 (iii) 0.7 (iv) 0.9 Choose the right answer from abo
The boiling point of 0.1 molal aqueous solution of urea is 100.18oC at 1 atm. The molal elevation constant of water is: (a) 1.8 (b) 0.18 (c) 18 (d) 18.6Answer: (a) Kb
Provide solution of this question. If 25 ml of 0.25 M NaCl solution is diluted with water to a volume of 500ml the new concentration of the solution is : (a) 0.167 M (b) 0.0125 M (c) 0.833 M (d) 0.0167 M
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
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