A 5 % solution ( by mass ) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5 % glucose in water if freezing point of pure water is 273.15 K.

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Solution

$5 %$ by mass of cane sugar means 100 g of solution has 5 g of cane sugar. The molecular weight of cane sugar is 342 g/mol. Number of moles of cane sugar $= \frac{5 g}{342 g / m o l} = 0.0146 m o l$
Mass of water $= 100 g - 5 g = 95 g = 0.095 k g$
$(∵ 1 k g = 1000 g)$
Molality $m = \frac{0.0146 m o l}{0.095 k g} = 0.154 m o l / k g$
The depression in the freezing point $Δ T_{f} = 273.15 K - 271 K = 2.15 K$
$Δ T_{f} = K_{f} m$
$K_{f} = \frac{Δ T_{f}}{m} = \frac{2.15 K}{0.154 m o l / K g} = 13.97 K K g / m o l$
$5 %$ by mass of glucose means 100 g of solution has 5 g of glucose. The molecular weight of glucose is 180 g/mol. Number of moles of glucose $= \frac{5 g}{180 g / m o l} = 0.0277 m o l$
Mass of water $= 100 g - 5 g = 95 g = 0.095 k g$
$(∵ 1 k g = 1000 g)$
Molality $m = \frac{0.0277 m o l}{0.095 k g} = 0.292 m o l / k g$
The depression in the freezing point
$Δ T_{f} = K_{f} m$
$Δ T_{f} = 0.292 m o l / 13.97 K K g / m o l \times 0.292 m o l / k g = 4.09 K$
The freezing point of glucose solution $= 273.15 K - 4.09 K = 269.07 K$

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A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

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