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December 19, 2014

December 19, 2014

Posted by **DrFunk** on Tuesday, June 16, 2009 at 4:56pm.

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 -
**GK**, Tuesday, June 16, 2009 at 8:13pm1.30 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 -
**ms**, Monday, August 2, 2010 at 2:04pm1.30 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

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