--%>
Assignment Help - homework help

Molality of glucose

Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal

   Related Questions in Chemistry

  • Q : Problem based on molarity Choose the

    Choose the right answer from following. The molarity of a solution of Na2CO3 having 10.6g/500ml of solution is : (a) 0.2M (b)2M (c)20M (d) 0.02M

    		•
  • Q : Molar concentration of hydrogen 20 g of

    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.

    		•
  • Q : What type of bond does HCl encompass

    What type of bond does HCl encompass? Describe briefly?

    		•
  • Q : Molarity of Barium hydroxide 25 ml of a

    25 ml of a solution of barium hydroxide on titration with 0.1 molar solution of the hydrochloric acid provide a litre value of 35 ml. The molarity of barium hydroxide solution will be: (i) 0.07 (ii) 0.14 (iii) 0.28 (iv) 0.35

    		•
  • Q : Problem on vapour pressure Choose the

    Choose the right answer from following. If P and P are the vapour pressure of a solvent and its solution respectively N1 and N2 and are the mole fractions of the solvent and solute respectively, then correct relation is: (a) P= PoN1 (b) P= Po N2 (c)P0= N2 (d)

    		•
  • Q : Basic concepts Determination of correct

    Determination of correct mol. Mass from Roult's law is applicable to :

    		•
  • Q : Calculating total number of moles

    Choose the right answer from following. While 90 gm of water is mixed with 300 gm of acetic acid. The total number of moles will be: (a)5 (b)10 (c)15 (d)20

    		•
  • Q : Pressure Phase Diagrams The occurrence

    The occurrence of different phases of a one component system can be shown on a pressure temperature. The phases present in a one line system at various temperatures can be conveniently presented on a P- versus-T diagram. An example is pro

    		•
  • Q : Explain Second Order Rate Equations.

    Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o

    		•
  • Q : Organic and inorganic substances living

    living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?

    		•