Question

The elevation of boiling point of water produced by dissolving 1.17 g sodium chloride in 100g water
(Kb = 0.512 K kg mol-') is
​

Answer 1

0.205 K

Explanation:

First of all we need to find out the molality of solution

Molality = number of moles of solute/ kg of solvent

Moles = mass / molar mass

Thus,

Molality= (1.17/58.44)/0.1kg= 0.2 mol per kg

Now as we know  that sodium chloride dissociates into two ions e.g. Na+ and Cl-

Hence, molality of ions = 2 x 0.2 = 0.4 moles per kg

Now,

Elevation of boiling = molality of ions x Kb

Elevation of boiling = 0.4 x 0.512= 0.205 K

Answer 2

Answer:

the answer is 0.205 k.

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