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A research chemist isolates a new compound with an
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Answer #1

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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