## Answers

3.

Depresion of freezing point is expressed as

delta T = K_{f} x molality

molality = delta T / K_{f}

Here delta T = 0.325 ^{o}C

K_{f} = 1.86 ^{o}C/m

So, molality = delta T / K_{f}

= 0.325 ^{o}C / (1.86 ^{o}C/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 = C_{3}H_{3}O_{2}.

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(C_{3}H_{3}O_{2}) = **C _{6}H_{6}O_{4}**

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