--%>
Assignment Help - homework help

Problem on molality

Select the right answer of the question. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g /ml : (a) 10.43 (b) 20.36 (c) 12.05 (d) 14.05

   Related Questions in Chemistry

  • Q : Determining highest normality What is

    What is the correct answer. Which of the given solutions contains highest normality: (i) 8 gm of KOH/litre (ii) N phosphoric acid (iii) 6 gm of NaOH /100 ml (iv) 0.5M H2SO4

    		•
  • Q : Which is largest planet in our solar

    which is largest planet in our solar system

    		•
  • Q : Henry law question Answer the following

    Answer the following qustion. The definition “The mass of a gas dissolved in a particular mass of a solvent at any temperature is proportional to the pressure of gas over the solvent” is: (i) Dalton’s Law of Parti

    		•
  • Q : Volume hydrogen peroxide Choose the

    Choose the right answer from following. The normality of 10 lit. volume hydrogen peroxide is: (a) 0.176 (b) 3.52 (c) 1.78 (d) 0.88 (e)17.8

    		•
  • Q : Molality of a glucose solution What

    What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m  (ii) 0.5m  (iii) 0.2m  (iv) 2m

    		•
  • Q : Which is polar HCl or HF Which one is

    Which one is polar HCl or HF?

    		•
  • Q : Solubility product On passing H 2 S gas

    On passing H2S gas through a particular solution of Cu+ and Zn+2 ions, first CuS is precipitated because : (a)Solubility product of CuS is equal to the ionic product of ZnS (b) Solubility product of CuS is equal to the solubility product

    		•
  • Q : Vapour pressure of benzene Give me

    Give me answer of this question. The vapour pressure of benzene at a certain temperature is 640mm of Hg. A non-volatile and non-electrolyte solid weighing 2.175g is added to 39.08g of benzene. The vapour pressure of the solution is 600,mm of Hg . What is the mo

    		•
  • 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 : Vapour pressure of the pure hydrocarbons

    Give me answer of this question. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be: (a) 0.549 (b)

    		•