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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Chemistry
Chapter
Solutions

JIPMER 2015 · Chemistry · Solutions

The molal freezing point depression constant for benzene (C_6 H_6)...

The molal freezing point depression constant for benzene (C6H6)\left(\mathrm{C}_6 \mathrm{H}_6\right) is 4.90 K kgmol−14.90 \mathrm{~K} \mathrm{~kg} \mathrm{mol}^{-1}. Selenium exists as a polymer of the type Sex\mathrm{Se}_x. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112∘C0.112^{\circ} \mathrm{C} lower than that of pure benzene. The molecular formula of selenium is (atomic mass of Se=78.8 g mol−1\mathrm{Se}=78.8 \mathrm{~g} \mathrm{~mol}^{-1})

Options

  1. ASe8\mathrm{Se}_8
  2. BSe6\mathrm{Se}_6
  3. CSe4\mathrm{Se}_4
  4. DSe2\mathrm{Se}_2

Correct answer

A. Se8\mathrm{Se}_8

Step-by-step solution

To calculate the molar mass of selenium (Se) MB=Kf×WBΔTf×WAM_B=\frac{K_f \times W_B}{\Delta T_f \times W_A} Mas of selenium (WB)=3.26 g\left(W_B\right)=3.26 \mathrm{~g} Mass of benzene (WA)=226 g=0.226 kg\left(W_A\right)=226 \mathrm{~g}=0.226 \mathrm{~kg} Depression in freezing point (ΔTf)=0.112∘C\left(\Delta T_f\right)=0.112^{\circ} \mathrm{C} Molar depression constant (Kf)=4.9 K kg mol−1∴MB=(4.9 K kg mol−1)(3.26 g)(0.112 K)(0.226 kg)=631.08 g mol−1\left(K_f\right)=4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \therefore M_B=\frac{\left(4.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)(3.26 \mathrm{~g})}{(0.112 \mathrm{~K})(0.226 \mathrm{~kg})}=631.08 \mathrm{~g} \mathrm{~mol}^{-1} Given, gram atomic mass of selenium =78.8 g/mol=78.8 \mathrm{~g} / \mathrm{mol} For Sex(78.8 g mol−1)=631.08 g mol−1\mathrm{Se}_x\left(78.8 \mathrm{~g} \mathrm{~mol}^{-1}\right)=631.08 \mathrm{~g} \mathrm{~mol}^{-1} or, x=631.08 g mol−178.8 g mol−1=8\quad x=\frac{631.08 \mathrm{~g} \mathrm{~mol}^{-1}}{78.8 \mathrm{~g} \mathrm{~mol}^{-1}}=8 Thus, molecular formula of selenium =Se8=\mathrm{Se}_8
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