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A 0.1539 molal aqueus solution of cane s...

A 0.1539 molal aqueus solution of cane sugar `("mol mass" = 342 g mol^(-1))` has a freezing point of 271 K while freezing point ofpure water is 273.15 K. What will be the freezing point of an aqueus solution containing 5 g of glucose `("mol. Mass" =180 g mol^(-1))` per 100 g of water?

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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?

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

A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271K. Calculate the freezing point of 5% aqueous glucdse solution if freezing point of water is 273.15K.

A 5% solution (w/W) of cane sugar (molar mass = 342 g mol^(-1) ) has freezing point 271 K. What will be the freezing point of 5% glucose (molar mass = 18 g mol^(-1) ) in water if freezing point of pure water is 273.15 K ?

A 5% solution (w/W) of cane sugar (molar mass = 342 g mol^(-1) ) has freezing point 271 K. What will be the freezing point of 5% glucose (molar mass = 18 g mol^(-1) ) in water if freezing point of pure water is 273.15 K ?