Assuming the boiling point increased by 2 â°c, what is the approximate molality of nacl when the video ends? kbp(water) = 0.512 â°c/(mol/kg)

Answer : The molality of NaCl is, 1.95 mol/kg

Explanation :

Formula used for Elevation in boiling point :

where,

= change in boiling point =

= boiling point constant  for water =

m = molality

i = Van't Hoff factor = 2 (for electrolyte)

The dissociation will be,

So, Van't Hoff factor = Number of solute particles = = 1 + 1 = 2

Now put all the given values in the above formula, we get:

Therefore, the molality of NaCl is, 1.95 mol/kg

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