--%>
Assignment Help - homework help

Neutralization of sodium hydroxide

How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g  (ii) 6 g (iii) 40 g  (iv) 60 g

E

Expert

Verified

The answer is (ii)

1500cm3 of .01 N HCL have number of gm equivalence

= N1 x V1/ 1000 = 1500x0.1/ 1000 = 0.15

0.15 gm. Equivalent of NaOH = 0.15 x 40 = 6 gm

   Related Questions in Chemistry

  • Q : Metallic chemistry why transation

    why transation metals show charaterstic colours to the flame?

    		•
  • Q : Statement of Henry law Determine the

    Determine the correct regarding Henry’s law: (1) The gas is in contact with the liquid must behave as an ideal gas (2) There must not be any chemical interaction among the gas and liquid (3) The pressure applied must be high (4) All of these.

    		•
  • Q : Finding Active mass of urea Can someone

    Can someone please help me in getting through this problem. 10 litre solution of urea comprises of 240 gm urea. The active mass of urea is: (i) 0.04 (ii) 0.02 (iii) 0.4 (iv) 0.2

    		•
  • Q : How to calculate solutions molar

    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

    		•
  • Q : What are isotonic and hypotonic

    The two solutions which are having equivalent osmotic pressure are called isotonic solutions. The isotonic solutions at the same temperature also have same molar concentration. If we have solutions having different osmotic pressures then the solution having different

    		•
  • Q : Question of vapour pressure Choose the

    Choose the right answer from following. Vapour pressure of a solution is: (a) Directly proportional to the mole fraction of the solvent (b) Inversely proportional to the mole fraction of the solute (c) Inversely proportional to the mole fraction of the solvent (d

    		•
  • Q : Problem based on lowering in vapour

    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

    		•
  • Q : Meaning of molality of a solution The

    The molality of a solution will be: (i) Number of moles of solute per 1000 ml of solvent (ii) Number of moles of solute per 1000 gm of solvent (iii) Number of moles of solute per 1000 ml of solution (iv) Number of gram equivalents of solute per 1000 m

    		•
  • Q : Sedimentation and Velocity The first

    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

    		•
  • Q : Concentration of Calcium carbonate Help

    Help me to go through this problem. 1000 gms aqueous solution of CaCO3 contains 10 gms of carbonate. Concentration of the solution is : (a)10 ppm (b)100 ppm (c)1000 ppm (d)10000 ppm

    		•