A 5% soln mass of cane sugar has freezing pt of 271k. Calculate the freezing pt of 5% glucose in water if freezing pt of pure water is 273.15k
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5% glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water. Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K.
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Answer:
5% glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water. Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K.
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