Question

Define molal boiling elevation constant. Calculate the boiling point of aqueous solution of glucose containing 18 g of glucose in 180 gram of water(Ky water =0.52 k kg mol-1)​

Answer 1

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Explanation:

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Answer 2

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Using the relation ΔTb =mKb

m=

$\frac{180}{1}$

=0.1

Hence, ΔTb=0.1×0.52=0.052 K

Hence, the boiling point of the solution will increase by 0.052 K

New boiling point =373.15+0.052=373.202 k

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