1.

An aqueous solution of glucose boils at 100.01^(@)C. The molal elevationconstant for water is 0.5 K kg mol^(-1). What is the number of glucose molecules in the solution containing 100 g of water ?

Answer»

Solution :`Delta T_(B)=K_(b)XX m , 0.01 = 0.5 xx m` which gives m = 0.02.
Thus, 1000 G of water contain glucose = 0.02 mole
`therefore` 100 g of water contain glucose = 0.002 mole `= 0.002 xx 6.02xx10^(23)` MOLECULES `= 1.204xx10^(21)` molecules.


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