JEE MainChemistryModeratePhysical Chemistry

When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of
benzene (Kf = 1.72 K kg mol-1), a freezing point depression of 2 K is
observed. The van't Hoff factor (i) is:
tc 20 g us¶FkkWbd vEy (C11H8O2) dks 50 g csathu eas ?kksyk ( (Kf
= 1.72 K kg mol-1) rc fgekad esa voueu 2 K A okUVgkWQ
xq.kkad (i) gksxk

Show answer and solution

\Delta T_{f} = i \times K_{f} \times \ molality\

(eksyyrk)

\Rightarrow 2 = i \times 1.72 \times \frac{20}{172} \times \frac{1000}{50}

\Rightarrow 2 = 4i

\Rightarrow i = 1/2 = 0.5

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx

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