chemistry

posted by .

How would I go about doing this question, can someone explain it step by step please, my answer I keep getting is 1.17m CaCl2, but the answer is 1.35m CaCl2?

A 1.30M solution of CaCl2 in water has a density of 1.1g/mL/ What is the MOLALITY?

Also, when for questions with relation to finding minimum pressure, or boiling point elevation or freezing point deperession quesitons, how do I know that I have to use the "i" factor or not?

Thanks

  • chemistry -

    1.30 M means 1.30 moles / liter of solution
    1 mole CaCl2 = 111.0 g (How do you get that?)
    1.30 mol x 111 g = 144.3 g CaCl2
    1.00 L = 1000 mLs
    volume x density = mass:
    1000 mLs x 1.1 g/mL = 1100 grams solution (total)
    1100 g - 144.3g = 955.7 g H2O ---> 0.9553 kg H2O
    molality = m = 1.3 mol CaCl2 / 0.9553 kg H2O = ???
    The above answer should be multiplied by 3 since 1 mole of CaCl2 produces THREE particles in solution. The new answer is the EFFECTIVE molality you use to get vapor pressure, melting point lowering , and boiling point rise. You got some studying to do on colligative properties and calculations related to that. We do not teach complete units on a bulletin board posting.

  • chemistry -

    1.30 M = 1.30mols of solute/L of solvent
    mass of Cacl2 = 1.30 mols CaCl2 x 111g/mol CaCl2(molar mass)= 144.3g of CaCl2 (solute)
    Solution = 1.11g/ml
    mass of solution = 1.11g/ml x 1000ml/L x1L
    = 1110g (solution)
    mass of solvent = 1110g (solution) - 144.3g (solute)= 965.7g of solvent x 1kg/1000g = 0.9657kg
    molality = 1.30 mols of solute/0.9657kg solvent
    = 1.35m

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry HELP!!!

    Only one question, it would be greatly appreciated if you would help me Thanx In an experiment, Na2CO3(aq) was added to 25.0 mL of CaCl2(aq) solution until no more precipitate was formed. The mass of precipitate produced was 2.50 g. …
  2. Chemistry

    CaCl2(s) Ca2+(aq) + 2 Cl-(aq) H = -81.5 kJ A 14.0 g sample of CaCl2 is dissolved in 130. g of water, with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming …
  3. chemistry

    A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution?
  4. chemistry

    If 2.1 moles of CaCl2 are dissolved in enough water to make a 0.92-liter solution, what is the resulting molarity?
  5. chemistry

    Calculate the solubility of Cacl2 in g/L based on the experiement. Given 650g of CaCl2 added to 1.0L and allowed to dissolve as much as possible. While the remaining solid CaCl2 is filtered out of the solution and found to have a mass …
  6. Chemistry

    If the molality of a CaCl2 solution is .100 mol/kg, and was made from 500.0 mL of water, how many grams of CaCl2 were needed to make the solution?
  7. Chemistry

    Dissolving solid CaCl2 H2O in a large volume of water is endothermic to the extent of 14.6 kj/mol. Also given: CaCl2(s)+6H2O(l)->CaCl2* 6H2O(s) delta H = -97.0 kj what is the heat of solution of anhydrous CaCl2 in a large volume …
  8. Chemistry

    Dissolving solid CaCl2 H2O in a large volume of water is endothermic to the extent of 14.6 kj/mol. Also given: CaCl2(s)+6H2O(l)->CaCl2* 6H2O(s) delta H = -97.0 kj what is the heat of solution of anhydrous CaCl2 in a large volume …
  9. Chemistry

    How mush CaCl2, in grams, is needed to make 2.0 L of a 3.5M solution?
  10. Chemistry

    CaCl2(s) -> Ca(aq) + 2Cl(aq) ΔH_rxn = +198.0 KJ "If that g sample of calcium chloride was dissolved in 1000.0g of water, with both substances originally at 25.0C, what would be the final temperature of the solution be?

More Similar Questions