Molarity based question
Help me to solve this problem. 4.0 gm of NaOH are contained in one decilitre of solution. Its molarity would be: (a) 4 M (b)2 M (c)1 M (d)1.5 M
The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea (b) BaCl2 > AlCl3 > Urea (c) Urea > BaCl2<
What do you mean by the term enzymes? Briefly illustrate it.
Answer the following qustion. The definition “The mass of a gas dissolved in a particular mass of a solvent at any temperature is proportional to the pressure of gas over the solvent” is: (i) Dalton’s Law of Parti
Select right answer of the question. A dry air is passed through the solution, containing the 10 gm of solute and 90 gm of water and then it pass through pure water. There is the depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. C
are halides are halogens more soluble? why?
Freezing point of a substance is the temperature at which solid and liquid phases of the substance coexist. It is defined as the temperature at which its solid and liquid phases have the same vapour pressure. The freezing point o
Illustrate a tripod? And how it’s used?
Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio
Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at 20°C for a solution containing 78g of benzene and 46g of toluene in torr is: (a) 50 (b)
Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<
18,76,764
1930282 Asked
3,689
Active Tutors
1422459
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!