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14/01/2022 3:18 pm
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40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is _____ K. (Nearest integer)
[Given : Kf = 1.86 K kg mol-1 ; Density of water = 1.00 g cm-3 ; Freezing point of water = 273.15 K]
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14/01/2022 3:23 pm
molality = \(\frac{(\frac{40}{180})mol}{0.2}Kg\)
= \(\frac{10}{9}\)molal
⇒ ΔTf = Tf - Tf' = 1.86 x \(\frac{10}{9}\)molal
⇒ Tf' = 273.15 - 1.86 x \(\frac{10}{9}\)
= 271.08 K ≃ 271 K (Nearest integer)