When 20g of naphthoic acid (c11h8o2) is dissolved in 50g benzene (kf = 1.72 kkgmol-1 ) a freezing point depression of 2k is observed. The vant hoff factor is?
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When 20g of naphthoic acid (C11,H8, O2) is dissolved in 50g of benzene (kf = 1.72K kg mol-1), a freezing point depression
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Correct option is
Answer::-0.5
Molaloity =
172×50/20×1000 =2.3
∴ Depression in freezing point =2=2.32×1.72×i
∴i=0.5
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