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ollegr [7]
2 years ago
5

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

Chemistry
1 answer:
muminat2 years ago
4 0

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

\Delta T_f = K_f \times m

Where,

\Delta T_f = Depression in freezing point

K_f = Molal depression constant

m = Molality

\Delta T_f = K_f \times m

1.33 = 5.12 \times m

m = 0.26

Molality = \frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}

Mass of solvent (toluene) = 15.0 g = 0.015 kg

0.26 = \frac{Mole\ of\ compound}{0.015}

Moles of compound = 0.015 × 0.26 = 0.00389 mol

Mol = \frac{Mass\ in\ g}{Molecular\ weight}

Mass of the compound = 1.450 g

Molecular\ weight = \frac{Mass\ in\ g}{Moles}

Molecular weight = \frac{1.450}{0.00389} = 372.13\ g/mol

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