# A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has a freezing point of 271 K. Calculate the freezing point of 5% aqueous glucose solution.

##### Given: Percentage by mass of cane sugar solution = 5 %Percentage by mass of glucose solution = 5 %,Freezing point of cane sugar solution = 271 KMolar mass of cane sugar = 342 g mol-1To find: Freezing point of glucose solutionFormula: ${\text{M}}_{2}=\frac{1000×{\text{K}}_{\text{f}}×{\text{W}}_{2}}{△{\text{T}}_{\text{f}}{\text{W}}_{1}}$Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W2) = 5 g, and mass of solvent (W1) = 95 g.5 % glucose solution means that mass of glucose = ${\text{W}}_{2}^{\prime }$ = 5g, and mass of solvent = ${\text{W}}_{1}^{\prime }$ = 95 gMolar mass of glucose (C6H12O6) = $\left({\text{M}}_{2}^{\prime }\right)$ = 180 g mol-1Δ Tf for cane sugar solution = $△{\text{T}}_{\text{f}}={\text{T}}_{\text{f}}^{0}-{\text{T}}_{\text{f}}$ = 273.15 K - 271 K = 215 KNow, using the formula,${\text{M}}_{2}=\frac{1000×{\text{K}}_{\text{f}}×{\text{W}}_{2}}{△{\text{T}}_{\text{f}}{\text{W}}_{1}}$Rearranging the formula, we get$1000{\text{K}}_{\text{f}}=\frac{{\text{M}}_{2}×△{\text{T}}_{\text{f}}×{\text{W}}_{1}}{{\text{W}}_{2}}$        ......(1)$1000{\text{K}}_{\text{f}}=\frac{{\text{M}}_{2}^{\prime }×△{\text{T}}_{\text{f}}^{\prime }×{\text{W}}_{1}^{\prime }}{{\text{W}}_{2}^{\prime }}$      .....(2)From equations (1) and (2),$\frac{{\text{M}}_{2}×△{\text{T}}_{\text{f}}×{\text{W}}_{1}}{{\text{W}}_{2}}=\frac{{\text{M}}_{2}^{\prime }×△{\text{T}}_{\text{f}}^{\prime }×{\text{W}}_{1}^{\prime }}{{\text{W}}_{2}^{\prime }}$∴ $\frac{342{\text{mol}}^{-1}×2.15\text{K}×95\text{g}}{\text{5 g}}=\frac{180{\text{g mol}}^{-1}×△{\text{T}}_{\text{f}}^{\prime }×95\text{g}}{\text{5 g}}$$△{\text{T}}_{\text{f}}^{\prime }=\frac{342{\text{mol}}^{-1}×2.15\text{K}}{180{\text{g mol}}^{-1}}$ = 4.085 K∴ Freezing point of glucose solution (Tf) = ${\text{T}}_{\text{f}}^{0}-△{\text{T}}_{\text{f}}^{\prime }$= 273.15 K - 4.085 K= 269.065 KFreezing point of glucose solution is 269.065 K.Alternate method:Formulae: 1. m = $\frac{1000{\text{W}}_{2}}{{\text{M}}_{2}{\text{W}}_{1}}$2. Δ Tf = KfmCalculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W2) = 5 g, and mass of solvent (W1) = 95 g.Now, using formula (i),Molality of cane sugar solution = m = $\frac{1000{\text{W}}_{2}}{{\text{M}}_{2}{\text{W}}_{1}}$$=\frac{1000{\text{g kg}}^{-1}×5\text{g}}{342{\text{g mol}}^{-1}×95\text{g}}$= 0.1539 mNow, Δ Tf for cane sugar solution = Δ Tf = $△{\text{T}}_{\text{f}}^{0}-{\text{T}}_{\text{f}}$ = 273.15 K - 271 K = 2.15 KFrom formula (ii),Δ Tf (cane sugar) = Kf × m${\text{K}}_{\text{f}}=\frac{△{\text{T}}_{\text{f}}}{\text{m}}$∴ ${\text{K}}_{\text{f}}=\frac{2.15}{0.1539}$ = 13.97 K kg mol-15 % glucose solution means that mass of glucose = ${\text{W}}_{2}^{\prime }$ = 5 g,and mass of solvent = ${\text{W}}_{1}^{\prime }$ = 95 gMolar mass of glucose (C6H12O6) = $\left({\text{M}}_{2}^{\prime }\right)$ = 180 g mol-1Using formula (i),Molality of glucose solution =m = $\frac{1000{\text{W}}_{2}}{{\text{M}}_{2}{\text{W}}_{1}}$= $\frac{1000{\text{g kg}}^{-1}×5\text{g}}{180{\text{g mol}}^{-1}×95\text{g}}$ = 0.2924 mFrom formula (ii),$△{\text{T}}_{\text{f}}^{\prime }\left(\text{glucose}\right)={\text{K}}_{\text{f}}×\text{m}$(⸪ Since solvent is same, Kf is same for cane sugar and glucose solutions.)∴ $△{\text{T}}_{\text{f}}^{\prime }\left(\text{glucose}\right)=13.97×0.2924=4.085\text{K}$∴ Freezing point of glucose solution $\left({\text{T}}_{\text{f}}\right)={\text{T}}_{\text{f}}^{0}-△{\text{T}}_{\text{f}}^{\prime }$= 273.15 K - 4.085 K= 269.065 K

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