Problem on Osmotic Pressure of solution
The osmotic pressure of a 5% solution of cane sugar at 150oC is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm
Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg (
Nucleophilic substitution reactions in halides containing - X bond may take place through either of the two different mechanisms,S<
Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules
The nuclear states produced by a magnetic field are studied in nuclear magnetic resonance spectroscopy.The frequency of the radiation that corresponds to the nuclear magnetic energy level spacings and the weakness of the radiation absorption that must be e
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
ive me answer of this question. When mercuric iodide is added to the aqueous solution of potassium iodide, the: (a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change
Which of the following solutions will have a lower vapour pressure and why? a) A 5% aqueous solution of cane sugar. b) A 5% aqueous solution of urea.
The normality of 10 lit. volume hydrogen peroxide is: (a) 0.176 (b) 3.52 (c) 1.78 (d) 0.88 (e)17.8
Select the right answer of the following question.What does not change on changing temperature : (a) Mole fraction (b) Normality (c) Molality (d) None of these
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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