An aqueous solution containing 17.5 g of an unknown molecular compound in 100.0 g of water was found to have a freezing point of -1.8
(change in freezing point) = (molal freezing point depression
constant)(effective molality of solution)
1.8C = (1.86 C/m)(molality)
molality = 0.9677
molality = (moles solute) / (kg solvent)
0.9677 = (moles solute) / 0.100 kg
moles solute = 0.09677
molar mass = 17.5g / 0.09677 mol = 180.8 g/mol
mass of solute (w2) = 17.5 g
weight of solvent (water)(w1) = 100g
Kf for water = 1.86K Kg/mol
M = 1000 x Kf x w2/ Tf x w1
= 1000 x 1.86 x 17.5 / 1.8 x 100 = 32550/180= 180.83 g/mol
An aqueous solution containing 17.5 g of an unknown molecular compound in 100.0 g of water...
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