What are the freezing point and boiling point of a solution prepared by dissolving 18.6g of...
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What are the freezing point and boiling point of a solutionprepared by dissolving 18.6g of CaCl2 in 200.0g ofwater? (kf = -1.86�/m and kb= 0.512�/m)?
What are the freezing point and boiling point of a solutionprepared by dissolving 18.6g of CaCl2 in 200.0g ofwater? (kf = -1.86�/m and kb= 0.512�/m)?
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Let us calculate moles of CaCl2
Molar mass of CaCl2 = 110.98 g/mol
moles of CaCl2 = 18.6 g CaCl2 x( 1 mole CaCl2/ 110.98g CaCl2)
                         = 0.168 moles
Then convert grams of water to kilograms of H2O
200g x ( 1kg / 10^3 g) = 0.200 kg
Calculate molality
molality = moles of solute/ kg of solvent
            = 0.168/0.200
             = 0.84 mol/kg
For boiling point we will use the relation,
dela Tb = i(Kb)(m)
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