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Ideal Gas Equation

An element X ...

Question

An element X(Atomic mass = 25) exists as $X_{4}$ in benzene. 51 g of saturated solution of X in benzene was added to 50.0 g of pure benzene. The resulting solution showed a depression of freezing point of 0.55 K. Calculate the solubility of X per 100 g of benzene. $(K_{f}$ for benzene = 5.5 K kg $m o l^{- 1})$ ___

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Solution

Let x g be the mass of element in 51.0 g of saturated solution.
Mass of benzene in 51.0 g of saturated solution = 51.0 – x g
Total mass of benzene containing x g of solute = 50 + 51 – x = (101 – x) g
$Δ T_{f} = \frac{1000 K_{f} W_{B}}{M_{B} W_{A}} = \frac{1000 \times 5.5 \times x}{4 \times 25 \times (101 - x)} = 0.55 (g i v e n) \Rightarrow x = 1.0 g$
Hence, solubility $= \frac{W_{B} \times 100}{W_{A}} = \frac{1}{(51 - 1)} \times 100 = 2.0 g$

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