When 10^{-3} M solution of glucose in water | vedclass

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(D) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f}$.For glucose (a non-electrolyte),$\Delta T_{f} = 0^{\circ} C -

Vedclass · Accepted Answer

$-0.0372$

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(D) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f}$.For glucose (a non-electrolyte),$\Delta T_{f} = 0^{\circ} C - (-0.0186^{\circ} C) = 0.0186^{\circ} C$.Since $\Delta T_{f} = i 	imes m 	imes K_{f}$ and both solutions have the same molarity $(10^{-3} \ M)$,their molality $(m)$ is also the same.For $NaCl$,the van't Hoff factor $i = 2$ (as $NaCl ightarrow Na^{+} + Cl^{-}$).Therefore,$\Delta T_{f}(NaCl) = i 	imes \Delta T_{f}(	ext{glucose}) = 2 	imes 0.0186^{\circ} C = 0.0372^{\circ} C$.The freezing point of the $NaCl$ solution is $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 0^{\circ} C - 0.0372^{\circ} C = -0.0372^{\circ} C$.

When $10^{-3} \ M$ solution of glucose in water freezes at $-0.0186^{\circ} C$,then at what temperature $10^{-3} \ M$ solution of $NaCl$ will freeze (in $^{\circ} C$)?

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