--%>

mcq

Give me answer of this question. The normality of 10% (weight/volume) acetic acid is: (a)1 N (b)10 N (c)1.7 N (d) 0.83 N

   Related Questions in Chemistry

  • Q : Basic concepts Determination of correct

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

  • Q : Liquid Vapour Free Energies The free

    The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter

  • Q : Problem on MM equation How to obtain

    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).

  • Q : Molar mass of solute The boiling point

    The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is: (a) 5.8g mol-1  (b)

  • Q : Numerical The volume of water to be

    The volume of water to be added to 100cm3 of 0.5 N N H2SO4 to get decinormal concentration is : (a) 400 cm3 (b) 500cm3 (c) 450cm3 (d)100cm3

  • Q : Lowering of vapour pressure Help me to

    Help me to go through this problem. Lowering of vapour pressure is highest for: (a) urea (b) 0.1 M glucose (c) 0.1M MgSo4 (d) 0.1M BaCl2

  • Q : HCl is polar or non-polar Can you

    Can you please illustrate that HCl is polar or non-polar? Briefly illustrate it.

  • Q : Problem based on molality of glucose

    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 : Anti-aromatic and the non-aromatic

    What is main difference among anti-aromatic and the non-aromatic compounds?

  • Q : Question 6 A student was analyzing an

    A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu