Question

One mole of ${\left( {{C_6}{H_5}} \right)_3}C$. $OH$ dissolved in $1000g$ of $100\% $ sulphuric acid lowers the freezing point of sulphuric acid twice as one mole of $C{H_3}OH$ shows in $1000g$ of $100\% $ sulphuric acid. Assuming that $C{H_3}OH$ is neither dissociated nor associated with sulphuric acid. ${\left( {{C_6}{H_5}} \right)_3}COH$ shows $100\% $ dissociation. If true enter $1$ else $0$.

Answer 1

Hint: The freezing point depends on the number of ions dissociated or on the concentrations of the particles when the compounds will dissociate in their aqueous solution. The greater the dissociation, the lower will be the freezing point.

Complete answer:
The required reaction is:
\[C{H_2}OH + {H_2}S{O_4} \rightleftharpoons C{H_3}O{H_2}^ + + HS{O_4}^ - \] (two ions dissociated)
As given in the question that the depression in freezing point is twice, so that the numbers of ions dissociated must also be twice.
Now, the second required reaction:
\[{({C_6}{H_5})_3}COH + 2{H_2}S{O_4} \rightleftharpoons {({C_6}{H_5})_3}{C^ + } + H3{O^ + } + 2HSO{4^ - }\] (four ions dissociated)
So, ${\left( {{C_6}{H_5}} \right)_3}COH$ shows $100\% $ dissociation.
So, the statement is true.
The freezing point is defined as the temperature at which a liquid becomes a solid or freezes. The freezing point depends on the number of ions dissociated or on the concentrations of the particles when the compounds will dissociate in their aqueous solution. The greater the dissociation, the lower will be the freezing point.

Additional information: Degree of dissociation is defined as the fraction of total number molecules that dissociate into simpler molecules or ions at a given time and the degree of association is defined as the fraction of the total number of molecules that combine together to give the formation of a bigger molecule at a given time.
For example, let say the initial moles of $N{H_3}$ (gas) is $Y$. Let $X$ moles of $N{H_3}$ dissociate at equilibrium. Now if $X$ moles dissociate and from $Y$ moles of $N{H_3}$, then, the degree of dissociation of $N{H_3}$ would be $\dfrac{X}{Y}$.

Note:
We should remember that the depression in freezing point is a colligative property. The colligative property explains the property of the solution, which depends on the ratio of the number of solute particles to the solvent molecules. Also, does not depend on the nature of the chemicals present.