The freezing point of a 0.01 M aqueous gluco | vedclass

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(B) The depression in freezing point $\Delta T_f$ is directly proportional to the molarity $M$ of the solution,i.e.,$\Delta T_f \propto M$.Given,for $

Vedclass · Accepted Answer

$-0.10$

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(B) The depression in freezing point $\Delta T_f$ is directly proportional to the molarity $M$ of the solution,i.e.,$\Delta T_f \propto M$.Given,for $M_1 = 0.01 \ M$,$\Delta T_{f,1} = 0.18 \ ^oC$ (since freezing point is $-0.18 \ ^oC$).When equal volumes $(V)$ of two solutions are mixed,the final concentration $M_{mix}$ is given by $M_{mix} = \frac{M_1 V + M_2 V}{V + V} = \frac{M_1 + M_2}{2}$.$M_{mix} = \frac{0.01 + 0.002}{2} = 0.006 \ M$.Using the proportionality $\frac{\Delta T_{f,2}}{\Delta T_{f,1}} = \frac{M_{mix}}{M_1}$,we get $\Delta T_{f,2} = \frac{0.006}{0.01} 	imes 0.18 = 0.6 	imes 0.18 = 0.108 \ ^oC$.Thus,the new freezing point is $-0.108 \ ^oC$,which is approximately $-0.10 \ ^oC$.

The freezing point of a $0.01 \ M$ aqueous glucose solution at $1 \ \text{atmosphere}$ is $-0.18 \ ^oC$. To it,an addition of an equal volume of $0.002 \ M$ glucose solution will produce a solution with a freezing point of nearly ........... $^oC$.

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