--%>
Assignment Help - homework help

Problem on molarity-normality-molality

Can someone please help me in getting through this problem. The solution ofAl2(SO4)3 d = 1.253gm/m comprise 22% salt by weight. The molarity, normality and molality of the solution is: (1) 0.805 M, 4.83 N, 0.825 M (2) 0.825 M, 48.3 N, 0.805 M (3) 4.83 M, 4.83 N, 4.83 M (4) None

   Related Questions in Chemistry

  • Q : What are methods of phenol preparation

    Phenol was initially obtained by fractional distillation of coal

    		•
  • Q : Problem on colligative properties

    Choose the right answer from following. The magnitude of colligative properties in all colloidal dispersions is : (a) Lowerthan solution (b)Higher than solution(c) Both (d) None

    		•
  • Q : Problem based on molarity Select the

    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

    		•
  • Q : Water under pressure problem-henry law

    Can someone help me in going through this problem. The statement “When 0.003 moles of a gas are dissolved in 900 gm of water under a pressure of 1 atm, 0.006 moles will be dissolved under the pressure of 2 atm", signfies: (a)

    		•
  • Q : Explain physical properties of

    . Boiling pointsThe boiling points of monohalogen derivatives of benzene, which are all liquids, follow the orderIodo > Bromo > ChloroThe boiling points of isomeric dihalobe

    		•
  • Q : Modes of concentration Which of the

    Which of the given modes of expressing concentration is fully independent of temperature: (1) Molarity (2) Molality (3) Formality (4) Normality Choose the right answer from above.

    		•
  • Q : Difference among hcl gas and hcl acid

    What is the basic difference among hcl gas and hcl acid? Briefly state the difference?

    		•
  • Q : Haloalkane how haloalkane can be

    how haloalkane can be prepared by refluxing alcohol with hydrohalic acids

    		•
  • Q : Calculation of molecular weight Provide

    Provide solution of this question. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of ace

    		•
  • Q : Dipole attractions-London dispersion

    Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?

    		•