Calculate The Freezing Point Of A Solution Containing 60g Of Glucose (Molar Mass =180 G Mol−1) In 250 G Of Water.

The number of moles of glucose =\( \frac{Mass of glucose}{Molecular weight of glucose} \) \(\Rightarrow \frac{60 g}{180 g / mol} \) \(\Rightarrow 0.333 mol \)

The molality of glucose solution =\(\frac{Number of moles of glucose}{Mass of solvent (in kg)} \) \(\Rightarrow \frac{0.333 mol}{250 g * \frac{1 kg}{1000 g}} \) \(\Rightarrow 1.333 mol / kg \)

The depression in the freezing point of glucose solution :

\(\Delta T_{f} = K_{f} m = 1.86 Kg / mol * 1.333 mol / kg = 2.48 K \)

The freezing point of pure water is\( 0^{\circ} C \)

The freezing point of a solution containing 60g of glucose in 250 g of water. =

\( 0 – 2.48 = -2.480^{\circ} C \)

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