--%>
Assignment Help - homework help

problem based on molarity

Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

   Related Questions in Chemistry

  • Q : Quastion of finding vapour pressure

    Vapour pressure of CCl425Degree C at is 143mm of Hg0.5gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl4 .Find the vapour pressure of the solution (Density of CCl4 = = 1.58g /cm2): (a)141.43mm (b)

    		•
  • Q : Volume of solution containing solute

    What volume of solution contains 0.1 mole of the solute: (a) 100ml (b) 125ml  (c) 500ml (d) 62.5ml Choose the right answer from above.

    		•
  • Q : Calculating number of moles from

    Choose the right answer from following. If 0.50 mol of CaCl2 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ca3 (PO2)2 which can be formed: (a) 0.70 (b) 0.50 (c) 0.20 (d) 0.10

    		•
  • Q : Relative lowering in vapour pressure of

    Give me answer of this question. "Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute". Above statement is: (a) Henry law (b) Dulong and Petit law (c) Raoult's law (d) Le-Chatelier's pri

    		•
  • Q : Describe the properties of the

    Briefly describe the properties of the carbohydrates?

    		•
  • Q : Question related to molarity Help me to

    Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M

    		•
  • Q : Mole fraction of benzene Choose the

    Choose the right answer from following. In a solution of 8.7g benzene C6H6 and 46.0 gm toluene ,(C6, H5, CH3) the mole fraction of benzene in this solution is: (a)1/6 (b)1/5 (c)1/2 (d)1/3

    		•
  • Q : Units of Measurement Unit of

    Unit of measurement- These are also some systems for units:      (1) C.G.S.

    		•
  • 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 : Amines why o-toluidine is a weaker base

    why o-toluidine is a weaker base than aniline?

    		•