Calculate the freezing point of a solution containing 60g of glucose (Molar mass = 180 g $M o l^{- 1})$ in 250 g of water.
$(k_{f} o f w a t e r = 1.86 K k g m o l^{- 1})$

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Solution

$T_{f} = ?, W_{A} = 60 g, M_{A} = 180 g, W_{B} = 250 g, K_{f} = 1.86 K k g m o l^{- 1}$

$Δ T_{f} = K_{f} m$

$T_{f}^{o} - T_{f} = K_{f} \frac{W_{A} \times 1000}{M_{A} \times W_{B}}$

$(0^{\circ} C) - T_{f} = 1.86 \times \frac{60 \times 1000}{180 \times 250}$

$T_{f} = - 1.86 \times \frac{4}{3}$

$T_{f} = - 0.62 \times 4$

$T_{f} = - {2.48}^{\circ} C$

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Question 2.32

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Calculate the freezing point of a solution containing 60g of glucose (Molar mass = 180 g Mol−1) in 250 g of water. (kf of water=1.86 K kg mol−1)

Tf=?,WA=60g,MA=180g,WB=250g,Kf=1.86K kg mol−1 ΔTf=Kfm Tof−Tf=KfWA×1000MA×WB (0∘C)−Tf=1.86×60×1000180×250 Tf=−1.86×43 Tf=−0.62×4 Tf=−2.48∘C

Calculate the freezing point of a solution containing 60g of glucose (Molar mass = 180 g $M o l^{- 1})$ in 250 g of water.
$(k_{f} o f w a t e r = 1.86 K k g m o l^{- 1})$