Question

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)

Answer 1

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