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A 0.1539 molal aqueous solution of cane sugar (molar mass = 342 g $mol^{−1}$) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. mass = 180 g $mol^{−1}$) per 100 g of solution?

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1 Answer

269.27K
Hence (A) is the correct answer.
answered Jun 12, 2014 by sreemathi.v
 
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