The freezing point of a 0.01 m aqueous gluco | vedclass

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(B) For the first solution: $\Delta T_f = 0 - (-0.18) = 0.18^\circ C$. Since $\Delta T_f = K_f 	imes m$,we have $0.18 = K_f 	imes 0.01$,so $K_f = 18

Vedclass · Accepted Answer

$-0.108$

Vedclass · Answer

(B) For the first solution: $\Delta T_f = 0 - (-0.18) = 0.18^\circ C$. Since $\Delta T_f = K_f 	imes m$,we have $0.18 = K_f 	imes 0.01$,so $K_f = 18$. When equal volumes of two solutions are mixed,the new molality $m_{mix}$ is the average of the two molalities: $m_{mix} = \frac{m_1 + m_2}{2} = \frac{0.01 + 0.002}{2} = 0.006 \ m$. The new depression in freezing point is $\Delta T_{f(new)} = K_f 	imes m_{mix} = 18 	imes 0.006 = 0.108^\circ C$. Therefore,the new freezing point is $0 - 0.108 = -0.108^\circ C$.

The freezing point of a $0.01 \ m$ aqueous glucose solution is $-0.18^\circ C$. If an equal volume of $0.002 \ m$ glucose solution is added to it,the freezing point of the resulting solution will be ...... $^\circ C$.

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