Show the calculation of the freezing point of a solution made by dissolving 12.6 grams of the electrolyte nano3 in 200 grams of water. kf for water is 1.86 and the fp of pure water is 0oc.

The freezing point of the solution is

Explanation:-

Depression in freezing point is given by:

= Depression in freezing point

i= vant hoff factor = 2 (for which dissociates to give two ions )

= freezing point constant =

m= molality

Weight of solvent (water)= 200 g = 0.2 kg

Molar mass of electrolyte =  85 g/mol

Mass of electrolyte added = 12.6 g

The freezing point of the solution is

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"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"