What is the freezing point of a solution of 1.43 g mgcl2 in 100 g of water? Kf for water is 1.86°c/m for water.

Change in T = i x m x kf. Where i is the number of particles the compound dissociates, m is the molality, and kf is the constant specific to water.  MM of MgCl2 = 95.21 g/mol

MgCl2 will form three atomsi = 31Mg + 2Cl

kf = 1.86

moles of MgCl2 = 1.43g/95.21 = 0.015

m = moles solute/ kg solvent = 0.015/.1kg = .15m

Change in T = (3)(.15)(1.86) = .838

This is the depression of the freezing point of water, it is negative because the water's new freezing point is its normal freezing point minus the change in T.

0 - .838 = -.838

Explanation:

"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"