CBSE Class 12 · 2002 · Delhi · SolutionsShort Answer

The molal freezing point depression constant of benzene (C

_{6}

_{6}

) is 4.90 K kg mol

^{- 1}

. Selenium exists as a polymer of the type Se

_{x}

. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower than that of pure benzene. Deduce the molecular formula of selenium. (Atomic mass of Se = 78.8 g mol

^{- 1}

)

Answer: Se₈

Solution

Molality from the depression:

m = \frac{Δ T _{f}}{K _{f}} = \frac{0.112}{4.90} = 0.02286 mol kg^{- 1}

Moles of Se

_{x}

in 226 g benzene:

0.02286 \times 0.226 = 5.166 \times 1 0^{- 3} mol

Molar mass:

M = \frac{3.26}{5.166 \times 1 0 ^{- 3}} = 631 g mol^{- 1}

Atoms per molecule:

x = \frac{631}{78.8} \approx 8

The molecular formula is S e_{8}

CBSE Class 12 Board · 2002 · Delhi · Subjective

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