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A 5% solution (by mass) of canesugar in ...

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

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A 0.1539 molal aqueous solution of canesugar (molar mass = 342g mol^-1) has a freezing point 271K. While the freezing point of pure water is 273.15k. What will be the freezing point of an aqueous solution, containing 5g of glucose (Mol mass = 180g mol^-1) per 100g of solution?

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A solution of sucrose (molecular mass 342 u) is prepared by dissolving 0.84 g in 100g of water at 298K. Calculate the freezing point of water in the solution. (K_f for water = 1.86 K kg mol^-1)

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What mass of NaCl (molar mass= 58.5g mol^-1) must be dissolved in 65g of water to lower the freezing point by 7.5^@C ? The freezing point depression constant, K_f for water is 1.86 K kg mol^-1 . Assume vant Hoff factor for NaCl is 1.87.

A solution of sucrose (molecular mass 342 u) is prepared by dissolving 6.84g in 100g of water at 298K. Calculate the boiling point of the solution. ( K_b for water, 0.52K mol^-1 )

0.01M solution of KCl and BaCl_2 are prepared in water. The freezing points of KCl is found to be -2^@C . What freezing point would you expect for BaCl_2 solution assuming both KCl and BaCl_2 to be completely ionised?

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A 0.561 m solution of an unknown electrolyte depresses the freezing point of water by 2.93^@C . What is Vant Hoff factor for this electrolyte? The freezing point depression constant (K_f) for water is 1.86^@C Kg mol^-1 .

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