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Depression in Freezing Point

The molal fre...

Question

The molal freezing point depresssion constant of benzene is $4.90 {Kkgmol}^{- 1}$ . Selenium exists as a polymer of the type $S e_{x}$ . When $3.26 g$ of selenium is dissolved in $226 g$ of benzene, the observed freezing point is $0.112^{\circ} C$ lower than pure benzene. What is $x$ ?

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Solution

Depression in freezing point is given by
$Δ T_{f} = K_{f} m$
$0.112 = 4.90 \times \frac{3.26 / M_{s o l u t e}}{226} \times 1000$
$M_{s o l u t e} = 631 g / m o l$
Selenium exists as $S e_{x}$ and the molar mass of Se is 78.9 g/mol
$∴ x = 631 / 78.9 = 7.99$
So, the value of $x$ is 8

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