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expeople1 [14]
3 years ago
9

Assuming the boiling point increased by 2 °C, what is the approximate molality of NaCl when the video ends? Kbp(water) = 0.512

°C/(mol/kg)
Chemistry
1 answer:
makkiz [27]3 years ago
6 0

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

Explanation :

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

where,

\Delta T_b = change in boiling point = 2^oC

k_b = boiling point constant  for water = 0.512^oC/m

m = molality

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

The dissociation NaCl will be,

NaCl\rightarrow Na^++Cl^{-}

So, Van't Hoff factor = Number of solute particles = Na^++Cl^{-} = 1 + 1 = 2

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

2^oC=2\times (0.512^oC/m)\times m

m=1.95mol/kg

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

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