One mole of C6H53C- OH dissolved in 1000g of 100

CH_2OH+H_2SO_4⇌ CH_3OH_2^++HSO_4^ (two ions)]Since #160;depression of #160;freezing point is twice, the number of ions must be twice.So, #160; ...

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Standard XII

Chemistry

Depression in Freezing Point

One mole of ...

Question

One mole of $(C_{6} H_{5})_{3} C \cdot O H$ dissolved in 1000g of 100% sulphuric acid lowers the freezing point of sulphuric acid twice as one mole of $C H_{3} O H$ shows in 1000 g of 100% sulphuric acid. Assuming that $C H_{3} O H$ is neither dissociated nor associated with sulphuric acid . $(C_{6} H_{5})_{3} C O H$ show 100% dissociation, If true enter 1 else enter 0.

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Solution

$C H_{2} O H + H_{2} S O_{4} ⇌ C H_{3} O H_{2}^{+} + H S O_{4}^{-}$ (two ions)]
Since depression of freezing point is twice, the number of ions must be twice.

So, $(C_{6} H_{5})_{3} C O H + 2 H_{2} S O_{4} ⇌ (C_{6} H_{5})_{3} C^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$ (four ions)

Hence, $(C_{6} H_{5})_{3} C O H$ shows 100% dissociation

Hence, the statement is correct.

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CH2OH+H2SO4⇌CH3OH+2+HSO−4 (two ions)]Since depression of freezing point is twice, the number of ions must be twice.So, (C6H5)3COH+2H2SO4⇌(C6H5)3C++H3O++2HSO−4 (four ions)Hence, (C6H5)3COH shows 100% dissociation Hence, the statement is correct.