3.
Depresion of freezing point is expressed as
delta T = Kf x molality
molality = delta T / Kf
Here delta T = 0.325 oC
Kf = 1.86 oC/m
So, molality = delta T / Kf
= 0.325 oC / (1.86 oC/m)
= 0.1747 m
= 0.1747 moles/Kg
we have 2.51 g of our new compound in 100 g of water.
So, 25.1 g / 1000 g of water = 25.1 g/Kg of water.
Hence,
0.1747 moles = 25.1 g
1 mole = 25.1 / 0.1747 = 143.7 g
Empirical Formula = C3H3O2.
So, empirical mass = (3 x 12) + (3 x 1) + (2 x 16)
= 71g
Which indicates there are 2 empirical units per molecular formula.
Hence the molecular formula is 2(C3H3O2) = C6H6O4
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